Chapter 7 Cubes and Cube Roots (Additional Questions)

A perfect cube is an integer that is the cube of another integer. In other words, a number $n$ is a perfect cube if there exists an integer $k$ such that $n = k^3$. To determine if a number is a perfect cube, we can find its prime factorization. If the exponents of all prime factors in the factorization are multiples of 3, then the number is a perfect cube.

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Let's examine each option:

(A) 243

Prime factorization of 243:

$\begin{array}{c|cc} 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $243 = 3 \times 3 \times 3 \times 3 \times 3 = 3^5$. The exponent of the prime factor 3 is 5, which is not a multiple of 3. Therefore, 243 is not a perfect cube.

(B) 512

Prime factorization of 512:

$\begin{array}{c|cc} 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

So, $512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^9$. The exponent of the prime factor 2 is 9, which is a multiple of 3 ($9 = 3 \times 3$). We can write $512 = (2^3)^3 = 8^3$. Therefore, 512 is a perfect cube.

(C) 625

Prime factorization of 625:

$\begin{array}{c|cc} 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, $625 = 5 \times 5 \times 5 \times 5 = 5^4$. The exponent of the prime factor 5 is 4, which is not a multiple of 3. Therefore, 625 is not a perfect cube.

(D) 10000

Prime factorization of 10000:

$10000 = 10^4 = (2 \times 5)^4 = 2^4 \times 5^4$.

The exponents of the prime factors 2 and 5 are both 4, which are not multiples of 3. Therefore, 10000 is not a perfect cube.

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Based on the prime factorization of each option, only 512 is a perfect cube.

The correct option is (B) 512.

To find the unit digit of the cube of a number, we only need to look at the unit digit of the original number and find the unit digit of its cube.

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The given number is 47.

The unit digit of 47 is 7.

Now, we need to find the cube of the unit digit, which is $7^3$.

$7^3 = 7 \times 7 \times 7$

$7 \times 7 = 49$

The unit digit of 49 is 9.

Now multiply the unit digit of 49 by 7 (the unit digit of the original number):

$9 \times 7 = 63$

The unit digit of 63 is 3.

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Alternatively, we can calculate the full cube of 7:

$7^3 = 343$

The unit digit of 343 is 3.

Therefore, the unit digit of the cube of 47 will be the same as the unit digit of $7^3$, which is 3.

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The correct option is (B) 3.

Let $n$ be a negative integer. This means $n < 0$.

The cube of $n$ is $n^3 = n \times n \times n$.

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We know that the product of two negative numbers is a positive number. So, $n \times n$ will be positive.

$(n) \times (n) = (\text{negative}) \times (\text{negative}) = (\text{positive})$

Now, we multiply this positive result by the original negative integer $n$.

$(n \times n) \times n = (\text{positive}) \times (\text{negative}) = (\text{negative})$

Therefore, the cube of a negative integer is always a negative number.

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Let's consider an example. Let $n = -3$.

The cube of -3 is $(-3)^3$.

$(-3)^3 = (-3) \times (-3) \times (-3)$

First, $(-3) \times (-3) = 9$ (a positive number).

Then, $9 \times (-3) = -27$ (a negative number).

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Thus, the cube of a negative integer is always Negative.

The correct option is (B) Negative.

To calculate the cube of a fraction $\left(\frac{a}{b}\right)^3$, we cube the numerator and the denominator separately: $\left(\frac{a}{b}\right)^3 = \frac{a^3}{b^3}$.

Also, the cube of a negative number is always negative, as $(-x)^3 = (-x) \times (-x) \times (-x) = x^2 \times (-x) = -x^3$.

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We need to find the value of $\left(-\frac{2}{3}\right)^3$.

Using the rules mentioned above, we can write:

$\left(-\frac{2}{3}\right)^3 = \frac{(-2)^3}{3^3}$

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Now, calculate the cube of the numerator and the denominator:

Numerator: $(-2)^3 = (-2) \times (-2) \times (-2) = 4 \times (-2) = -8$

Denominator: $3^3 = 3 \times 3 \times 3 = 9 \times 3 = 27$

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So, $\left(-\frac{2}{3}\right)^3 = \frac{-8}{27}$.

This can also be written as $-\frac{8}{27}$.

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The value of $\left(-\frac{2}{3}\right)^3$ is $-\frac{8}{27}$.

The correct option is (B) $-\frac{8}{27}$.

A perfect cube is an integer that can be expressed as the cube of another integer. To determine if a number is a perfect cube, we can find its prime factorization or identify if it is the result of cubing an integer.

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Let's examine each option:

(A) 125

We can find the prime factorization of 125:

$\begin{array}{c|cc} 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, $125 = 5 \times 5 \times 5 = 5^3$. Since 125 is the cube of the integer 5, it is a perfect cube.

(B) 216

We can find the prime factorization of 216:

$\begin{array}{c|cc} 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^3 \times 3^3 = (2 \times 3)^3 = 6^3$. Since 216 is the cube of the integer 6, it is a perfect cube.

(C) 729

We can find the prime factorization of 729:

$\begin{array}{c|cc} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6 = (3^2)^3 = 9^3$. Since 729 is the cube of the integer 9, it is a perfect cube.

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Since 125, 216, and 729 are all perfect cubes, the correct answer is that all of the given numbers are perfect cubes.

The correct option is (D) All of the above.

To find the smallest number by which 108 must be multiplied to get a perfect cube, we first need to find the prime factorization of 108. A number is a perfect cube if in its prime factorization, the exponent of each prime factor is a multiple of 3.

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Prime factorization of 108:

$\begin{array}{c|cc} 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $108 = 2 \times 2 \times 3 \times 3 \times 3 = 2^2 \times 3^3$.

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In the prime factorization $2^2 \times 3^3$, the exponent of 3 is 3, which is already a multiple of 3. However, the exponent of 2 is 2, which is not a multiple of 3.

To make the exponent of 2 a multiple of 3 (the smallest multiple of 3 greater than 2 is 3), we need to multiply by $2^{3-2} = 2^1 = 2$.

So, if we multiply 108 by 2, the product will be:

$108 \times 2 = (2^2 \times 3^3) \times 2^1 = 2^{2+1} \times 3^3 = 2^3 \times 3^3 = (2 \times 3)^3 = 6^3 = 216$.

216 is a perfect cube ($6^3$).

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The smallest number by which 108 must be multiplied to make it a perfect cube is 2.

The correct option is (A) 2.

To find the smallest number by which 675 must be divided to get a perfect cube, we first need to find the prime factorization of 675. A number is a perfect cube if in its prime factorization, the exponent of each prime factor is a multiple of 3.

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Prime factorization of 675:

$\begin{array}{c|cc} 3 & 675 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, $675 = 3 \times 3 \times 3 \times 5 \times 5 = 3^3 \times 5^2$.

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In the prime factorization $3^3 \times 5^2$, the exponent of 3 is 3, which is a multiple of 3. However, the exponent of 5 is 2, which is not a multiple of 3.

To make the exponent of 5 a multiple of 3 (the largest multiple of 3 less than or equal to 2 is 0, which means eliminating the factor), we need to divide by $5^2 = 5 \times 5 = 25$.

So, if we divide 675 by 25, the quotient will be:

$\frac{675}{25} = \frac{3^3 \times 5^2}{5^2} = 3^3$.

$3^3 = 27$. 27 is a perfect cube ($3^3$).

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The smallest number by which 675 must be divided to make it a perfect cube is 25.

The correct option is (D) 25.

To find the cube root of 3375 by prime factorization, we follow these steps:

1. Find the prime factorization of 3375.

2. Group the prime factors into triplets.

3. For each triplet, take one factor.

4. Multiply the selected factors to find the cube root.

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Prime factorization of 3375:

$\begin{array}{c|cc} 3 & 3375 \\ \hline 3 & 1125 \\ \hline 3 & 375 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, the prime factorization of 3375 is $3 \times 3 \times 3 \times 5 \times 5 \times 5$.

We can write this in exponential form as $3^3 \times 5^3$.

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Now, we group the prime factors into triplets:

$3375 = (3 \times 3 \times 3) \times (5 \times 5 \times 5) = 3^3 \times 5^3$.

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To find the cube root $\sqrt[3]{3375}$, we take one factor from each triplet:

$\sqrt[3]{3375} = \sqrt[3]{3^3 \times 5^3} = \sqrt[3]{(3 \times 5)^3}$

Using the property $\sqrt[3]{a^3} = a$, we get:

$\sqrt[3]{3375} = 3 \times 5 = 15$.

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Thus, the cube root of 3375 is 15.

The correct option is (A) 15.

We need to find the cube root of $-125$. The cube root of a number $x$, denoted as $\sqrt[3]{x}$, is a number $y$ such that $y^3 = x$.

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We are looking for a number $y$ such that $y^3 = -125$.

We know that the cube of a positive number is positive, the cube of zero is zero, and the cube of a negative number is negative.

Since the number inside the cube root is negative ($-125$), the cube root must also be a negative number.

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Let's consider the absolute value of $-125$, which is $125$.

We know that $5 \times 5 \times 5 = 125$. So, the cube root of 125 is 5, i.e., $\sqrt[3]{125} = 5$.

Now, considering the negative sign, we can check if $(-5)^3$ equals $-125$.

$(-5)^3 = (-5) \times (-5) \times (-5)$

$(-5) \times (-5) = 25$

$25 \times (-5) = -125$

So, $(-5)^3 = -125$.

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Therefore, the cube root of $-125$ is $-5$.

$\sqrt[3]{-125} = -5$

The correct option is (B) $-5$.

To find the cube root of a fraction, we find the cube root of the numerator and the cube root of the denominator separately.

The property used is $\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}$.

In this case, we need to find $\sqrt[3]{\frac{64}{125}}$.

So, $\sqrt[3]{\frac{64}{125}} = \frac{\sqrt[3]{64}}{\sqrt[3]{125}}$.

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First, let's find the cube root of the numerator, 64. We can use prime factorization.

Prime factorization of 64:

$\begin{array}{c|cc} 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

So, $64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6$.

To find the cube root, we group the prime factors into triplets: $64 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) = 2^3 \times 2^3$.

The cube root of 64 is obtained by taking one factor from each triplet: $\sqrt[3]{64} = 2 \times 2 = 4$.

Alternatively, we know that $4^3 = 4 \times 4 \times 4 = 64$. So, $\sqrt[3]{64} = 4$.

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Next, let's find the cube root of the denominator, 125. We can use prime factorization.

Prime factorization of 125:

$\begin{array}{c|cc} 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, $125 = 5 \times 5 \times 5 = 5^3$.

To find the cube root, we group the prime factors into triplets: $125 = (5 \times 5 \times 5) = 5^3$.

The cube root of 125 is obtained by taking one factor from the triplet: $\sqrt[3]{125} = 5$.

Alternatively, we know that $5^3 = 5 \times 5 \times 5 = 125$. So, $\sqrt[3]{125} = 5$.

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Now, we combine the cube roots of the numerator and the denominator:

$\sqrt[3]{\frac{64}{125}} = \frac{\sqrt[3]{64}}{\sqrt[3]{125}} = \frac{4}{5}$.

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The cube root of $\frac{64}{125}$ is $\frac{4}{5}$.

The correct option is (A) $\frac{4}{5}$.

To find the cube root of a decimal number, we can first convert the decimal into a fraction. The number of decimal places in the original number helps determine the denominator.

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The given number is $0.008$. It has 3 decimal places.

We can write $0.008$ as a fraction:

$0.008 = \frac{8}{1000}$

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Now, we need to find the cube root of this fraction:

$\sqrt[3]{0.008} = \sqrt[3]{\frac{8}{1000}}$

Using the property $\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}$, we get:

$\sqrt[3]{\frac{8}{1000}} = \frac{\sqrt[3]{8}}{\sqrt[3]{1000}}$

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We find the cube root of the numerator and the denominator separately:

$\sqrt[3]{8}$: We look for a number that, when multiplied by itself three times, gives 8. We know that $2 \times 2 \times 2 = 8$. So, $\sqrt[3]{8} = 2$.

$\sqrt[3]{1000}$: We look for a number that, when multiplied by itself three times, gives 1000. We know that $10 \times 10 \times 10 = 1000$. So, $\sqrt[3]{1000} = 10$.

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Now, substitute these values back into the fraction:

$\frac{\sqrt[3]{8}}{\sqrt[3]{1000}} = \frac{2}{10}$

Convert the fraction back to a decimal:

$\frac{2}{10} = 0.2$

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Alternatively, for cube roots of decimals, count the number of decimal places. If it's a multiple of 3 (like 3 in 0.008), the cube root will have that number of decimal places divided by 3 ($3/3 = 1$). Find the cube root of the number ignoring the decimal point ($\sqrt[3]{8} = 2$) and place the decimal point accordingly (0.2).

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The cube root of $0.008$ is $0.2$.

The correct option is (A) 0.2.

To find the cube root of 17576 by estimation, we follow these steps:

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Step 1: Group the digits.

Group the digits of the number from the right in triplets. The two groups are 17 and 576.

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Step 2: Find the unit digit of the cube root.

Consider the rightmost group: 576.

The unit digit of this group is 6.

Now, look at the unit digits of the cubes of numbers from 1 to 9:

$1^3 = 1$

$2^3 = 8$

$3^3 = 27$ (unit digit 7)

$4^3 = 64$ (unit digit 4)

$5^3 = 125$ (unit digit 5)

$6^3 = 216$ (unit digit 6)

$7^3 = 343$ (unit digit 3)

$8^3 = 512$ (unit digit 2)

$9^3 = 729$ (unit digit 9)

The only digit whose cube ends in 6 is 6 ($6^3 = 216$).

So, the unit digit of the cube root of 17576 is 6.

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Step 3: Find the tens digit of the cube root.

Consider the leftmost group: 17.

Find two consecutive perfect cubes between which 17 lies.

$2^3 = 8$

$3^3 = 27$

We see that $8 < 17 < 27$.

This means $\sqrt[3]{8} < \sqrt[3]{17} < \sqrt[3]{27}$, which simplifies to $2 < \sqrt[3]{17} < 3$.

The tens digit of the cube root is the smaller of the two numbers whose cubes the group lies between. The smaller number is 2.

So, the tens digit of the cube root of 17576 is 2.

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Step 4: Combine the digits.

Combining the tens digit (2) and the unit digit (6), the estimated cube root of 17576 is 26.

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Verification:

Let's check if $26^3 = 17576$.

$26^3 = 26 \times 26 \times 26$

$26 \times 26 = 676$

Now, $676 \times 26$:

$\begin{array}{cc}& & 6 & 7 & 6 \\ \times & & & 2 & 6 \\ \hline && 4 & 0 & 5 & 6 \\ 1 & 3 & 5 & 2 & \times \\ \hline 1 & 7 & 5 & 7 & 6 \\ \hline \end{array}$

$26^3 = 17576$. The estimation is correct.

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By estimation, the cube root of 17576 is 26.

The correct option is (B) 26.

An even number is any integer that is divisible by 2. An even number can be expressed in the form $2k$, where $k$ is an integer.

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Let the even number be $n$. Then $n = 2k$ for some integer $k$.

We want to find the cube of this number, which is $n^3$.

Substitute $n = 2k$ into the expression for the cube:

$n^3 = (2k)^3$

Using the property of exponents $(ab)^m = a^m b^m$, we get:

$n^3 = 2^3 \times k^3$

Calculate $2^3$:

$2^3 = 2 \times 2 \times 2 = 8$

So, $n^3 = 8k^3$.

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We can rewrite $8k^3$ as $2 \times (4k^3)$.

Since $k$ is an integer, $k^3$ is also an integer, and therefore $4k^3$ is an integer.

Let $m = 4k^3$. Then $n^3 = 2m$, where $m$ is an integer.

By the definition of an even number, any integer that can be expressed in the form $2m$, where $m$ is an integer, is an even number.

Thus, the cube of an even number is always an even number.

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Let's check with some examples of even numbers:

If the even number is 2, its cube is $2^3 = 8$. 8 is an even number.

If the even number is 4, its cube is $4^3 = 64$. 64 is an even number.

If the even number is 6, its cube is $6^3 = 216$. 216 is an even number.

If the even number is -2, its cube is $(-2)^3 = -8$. -8 is an even number.

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The cube of an even number is always Even.

The correct option is (B) Even.

The given pattern shows that the cube of a natural number $n$ is the sum of $n$ consecutive odd numbers.

For $n=1$: Sum of 1 odd number = $1 = 1^3$. The first odd number is 1.

For $n=2$: Sum of 2 consecutive odd numbers = $3+5 = 8 = 2^3$. The consecutive odd numbers start from 3.

For $n=3$: Sum of 3 consecutive odd numbers = $7+9+11 = 27 = 3^3$. The consecutive odd numbers start from 7.

For $n=4$: Sum of 4 consecutive odd numbers = $13+15+17+19 = 64 = 4^3$. The consecutive odd numbers start from 13.

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We need to find the sum of the next set of consecutive odd numbers that equals $5^3$.

First, calculate $5^3$: $5^3 = 5 \times 5 \times 5 = 125$.

According to the pattern, $5^3$ will be the sum of 5 consecutive odd numbers.

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Let's observe the starting odd number for each value of $n$:

• $n=1$: starts at 1
• $n=2$: starts at 3 (Difference from previous start: $3-1=2$)
• $n=3$: starts at 7 (Difference from previous start: $7-3=4$)
• $n=4$: starts at 13 (Difference from previous start: $13-7=6$)

The difference between the starting odd numbers for $n$ and $n-1$ is $2(n-1)$.

For $n=5$, the difference from the starting number for $n=4$ should be $2(5-1) = 2 \times 4 = 8$.

The starting odd number for $n=5$ is the starting number for $n=4$ plus 8.

Starting number for $n=5 = 13 + 8 = 21$.

So, $5^3$ is the sum of 5 consecutive odd numbers starting from 21.

The 5 consecutive odd numbers starting from 21 are: 21, 23, 25, 27, 29.

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Let's verify the sum of these numbers:

$21 + 23 + 25 + 27 + 29$

Using addition:

$\begin{array}{cc} & 2 & 1 \\ & 2 & 3 \\ & 2 & 5 \\ & 2 & 7 \\ + & 2 & 9 \\ \hline 1 & 2 & 5 \\ \hline \end{array}$

The sum is 125, which is equal to $5^3$.

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The sum of the next set of consecutive odd numbers that equals $5^3$ is $21+23+25+27+29$.

The correct option is (A) $21+23+25+27+29$.

A perfect cube is an integer that can be expressed as the cube of another integer. To determine if a number is a perfect cube, we can find its prime factorization. If the exponent of each prime factor in the factorization is a multiple of 3, then the number is a perfect cube.

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Let's examine each option:

(A) 1331

Prime factorization of 1331:

$\begin{array}{c|cc} 11 & 1331 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

So, $1331 = 11 \times 11 \times 11 = 11^3$. Since the exponent of the prime factor 11 is 3 (a multiple of 3), 1331 is a perfect cube.

(B) 2744

Prime factorization of 2744:

$\begin{array}{c|cc} 2 & 2744 \\ \hline 2 & 1372 \\ \hline 2 & 686 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

So, $2744 = 2 \times 2 \times 2 \times 7 \times 7 \times 7 = 2^3 \times 7^3 = (2 \times 7)^3 = 14^3$. Since the exponents of the prime factors 2 and 7 are both 3 (multiples of 3), 2744 is a perfect cube.

(C) 4096

Prime factorization of 4096:

$4096 = 2 \times 2048 = 2^2 \times 1024 = 2^3 \times 512 = 2^3 \times 8^3 = 2^3 \times (2^3)^3 = 2^3 \times 2^9 = 2^{12}$.

Alternatively:

$\begin{array}{c|cc} 2 & 4096 \\ \hline 2 & 2048 \\ \hline 2 & 1024 \\ \hline 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

So, $4096 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{12}$. Since the exponent of the prime factor 2 is 12 (a multiple of 3, as $12 = 3 \times 4$), 4096 is a perfect cube ($4096 = (2^4)^3 = 16^3$).

(D) 5831

Prime factorization of 5831:

Try dividing by small prime numbers: not divisible by 2, 3 (sum of digits $5+8+3+1 = 17$), 5.

Try 7: $5831 \div 7 = 833$.

Try 7 again: $833 \div 7 = 119$.

Try 7 again: $119 \div 7 = 17$.

17 is a prime number.

$\begin{array}{c|cc} 7 & 5831 \\ \hline 7 & 833 \\ \hline 7 & 119 \\ \hline 17 & 17 \\ \hline & 1 \end{array}$

So, $5831 = 7 \times 7 \times 17 = 7^2 \times 17^1$. The exponents of the prime factors 7 and 17 are 2 and 1, respectively, which are not multiples of 3. Therefore, 5831 is NOT a perfect cube.

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Based on the prime factorization of each option, 5831 is the only number that is not a perfect cube.

The correct option is (D) 5831.

Let's analyze each statement:

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(A) A number ending in 7 always has its cube ending in 3.

The unit digit of the cube of a number is determined by the unit digit of the original number.

If a number ends in 7, its unit digit is 7. The unit digit of its cube will be the unit digit of $7^3$.

$7^1 = 7$

$7^2 = 49$ (Unit digit is 9)

$7^3 = 343$ (Unit digit is 3)

Since the unit digit of $7^3$ is 3, the cube of any number ending in 7 will end in 3.

This statement is TRUE.

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(B) A number ending in 0 has its cube ending in at least three zeroes.

If a number ends in 0, it is a multiple of 10. Let the number be $10n$ for some integer $n$.

Its cube is $(10n)^3 = 10^3 \times n^3 = 1000 \times n^3$.

Any number multiplied by 1000 will have at least three zeroes at the end.

For example, $10^3 = 1000$ (3 zeroes), $20^3 = 8000$ (3 zeroes), $100^3 = 1000000$ (6 zeroes).

If a number ends in $k$ zeroes, it is of the form $m \times 10^k$ where $m$ is not divisible by 10. Its cube is $(m \times 10^k)^3 = m^3 \times 10^{3k}$. The number of zeroes at the end is $3k$. Since the number ends in at least one zero ($k \ge 1$), the cube ends in at least $3 \times 1 = 3$ zeroes.

This statement is TRUE.

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(C) The cube of a number ending in 9 ends in 9.

If a number ends in 9, its unit digit is 9. The unit digit of its cube will be the unit digit of $9^3$.

$9^1 = 9$

$9^2 = 81$ (Unit digit is 1)

$9^3 = 729$ (Unit digit is 9)

Since the unit digit of $9^3$ is 9, the cube of any number ending in 9 will end in 9.

This statement is TRUE.

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(D) There is no perfect cube ending in 2 or 8.

Let's check the unit digits of cubes:

• $1^3$ ends in 1
• $2^3 = 8$ (ends in 8)
• $3^3$ ends in 7
• $4^3$ ends in 4
• $5^3$ ends in 5
• $6^3$ ends in 6
• $7^3$ ends in 3
• $8^3 = 512$ (ends in 2)
• $9^3$ ends in 9
• $10^3$ ends in 0

Perfect cubes ending in 8 exist (e.g., $2^3 = 8$, $12^3 = 1728$).

Perfect cubes ending in 2 exist (e.g., $8^3 = 512$, $18^3 = 5832$).

This statement is FALSE.

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The statements that are TRUE are (A), (B), and (C).

Let's evaluate the Assertion (A) and the Reason (R).

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Assertion (A): The cube of a single-digit number can be a single-digit number.

Single-digit numbers are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

Let's find the cubes of the first few single-digit numbers:

$0^3 = 0$ (a single-digit number)

$1^3 = 1$ (a single-digit number)

$2^3 = 8$ (a single-digit number)

$3^3 = 27$ (a two-digit number)

Since the cubes of 0, 1, and 2 are single-digit numbers, the assertion that the cube of a single-digit number *can be* a single-digit number is true.

Assertion (A) is TRUE.

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Reason (R): $1^3 = 1$ and $2^3 = 8$.

This statement provides specific calculations for the cubes of 1 and 2.

The calculation $1^3 = 1 \times 1 \times 1 = 1$ is correct.

The calculation $2^3 = 2 \times 2 \times 2 = 8$ is correct.

Reason (R) is TRUE.

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Now, let's consider if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) claims that it is possible for a single-digit number's cube to be a single-digit number. Reason (R) provides two examples ($1^3=1$ and $2^3=8$) where this possibility is demonstrated. Both 1 and 2 are single-digit numbers, and their cubes (1 and 8) are also single-digit numbers.

These examples directly support and explain why Assertion (A) is true.

Therefore, Reason (R) is the correct explanation of Assertion (A).

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Based on the analysis, both A and R are true, and R is the correct explanation of A.

The correct option is (A) Both A and R are true, and R is the correct explanation of A.

We need to find the cube root for each number in the first column and match it with the corresponding number in the second column. We can do this by cubing the numbers in the second column or finding the cube roots of the numbers in the first column.

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Let's calculate the cubes of the numbers in column (a), (b), (c), (d):

(b) 11: $11^3 = 11 \times 11 \times 11 = 121 \times 11 = 1331$.

So, the cube root of 1331 is 11. This matches (i) with (b).

(d) 13: $13^3 = 13 \times 13 \times 13 = 169 \times 13$.

$\begin{array}{cc}& & 1 & 6 & 9 \\ \times & & & 1 & 3 \\ \hline && 5 & 0 & 7 \\ 1 & 6 & 9 & \times \\ \hline 2 & 1 & 9 & 7 \\ \hline \end{array}$

So, $13^3 = 2197$. The cube root of 2197 is 13. This matches (ii) with (d).

(c) 17: $17^3 = 17 \times 17 \times 17 = 289 \times 17$.

$\begin{array}{cc}& & 2 & 8 & 9 \\ \times & & & 1 & 7 \\ \hline & 2 & 0 & 2 & 3 \\ 2 & 8 & 9 & \times \\ \hline 4 & 9 & 1 & 3 \\ \hline \end{array}$

So, $17^3 = 4913$. The cube root of 4913 is 17. This matches (iii) with (c).

(a) 19: $19^3 = 19 \times 19 \times 19 = 361 \times 19$.

$\begin{array}{cc}& & 3 & 6 & 1 \\ \times & & & 1 & 9 \\ \hline & 3 & 2 & 4 & 9 \\ 3 & 6 & 1 & \times \\ \hline 6 & 8 & 5 & 9 \\ \hline \end{array}$

So, $19^3 = 6859$. The cube root of 6859 is 19. This matches (iv) with (a).

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The correct matches are:

• (i) 1331 - (b) 11
• (ii) 2197 - (d) 13
• (iii) 4913 - (c) 17
• (iv) 6859 - (a) 19

Comparing these matches with the given options, we find that option (A) is correct.

The correct option is (A) (i)-(b), (ii)-(d), (iii)-(c), (iv)-(a).

We need to find the cube root of the product $27 \times 64$. We can use the property of cube roots that $\sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b}$.

So, $\sqrt[3]{27 \times 64} = \sqrt[3]{27} \times \sqrt[3]{64}$.

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First, find the cube root of 27:

We know that $3 \times 3 \times 3 = 3^3 = 27$.

So, $\sqrt[3]{27} = 3$.

---

Next, find the cube root of 64:

We know that $4 \times 4 \times 4 = 4^3 = 64$.

So, $\sqrt[3]{64} = 4$.

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Now, multiply the cube roots:

$\sqrt[3]{27} \times \sqrt[3]{64} = 3 \times 4 = 12$.

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Alternatively, we could first multiply 27 and 64, and then find the cube root of the product.

$27 \times 64$:

$\begin{array}{cc}& & 2 & 7 \\ \times & & 6 & 4 \\ \hline & 1 & 0 & 8 \\ 1 & 6 & 2 & \times \\ \hline 1 & 7 & 2 & 8 \\ \hline \end{array}$

$27 \times 64 = 1728$.

Now, find the cube root of 1728. We can use prime factorization or estimation.

Using prime factorization:

$\begin{array}{c|cc} 2 & 1728 \\ \hline 2 & 864 \\ \hline 2 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^6 \times 3^3$.

$\sqrt[3]{1728} = \sqrt[3]{2^6 \times 3^3} = \sqrt[3]{(2^2)^3 \times 3^3} = \sqrt[3]{4^3 \times 3^3} = \sqrt[3]{(4 \times 3)^3} = 4 \times 3 = 12$.

Using estimation:

Group the digits of 1728 into 1 and 728.

The unit digit of 728 is 8. Since $2^3$ ends in 8, the unit digit of the cube root is 2.

The leftmost group is 1. The perfect cube less than or equal to 1 is $1^3 = 1$. So the tens digit is 1.

The estimated cube root is 12. Check: $12^3 = 1728$.

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Both methods give the cube root as 12.

The correct option is (A) $3 \times 4 = 12$.

The case study provides information about the arrangement of laddoos in a cubical box.

---

The problem statement explicitly states: "A sweet shop owner in Kolkata wants to arrange 9261 laddoos in layers...".

This sentence directly provides the total number of laddoos.

---

The subsequent information confirms that this total number (9261) is a perfect cube, and its cube root gives the number of layers and the number of laddoos in each layer. This is consistent with a cubical arrangement where the total number of items is the cube of the number of items along one edge (or layer/dimension).

Total laddoos = (Number of laddoos in a layer) $\times$ (Number of layers)

Since the number of laddoos in each layer is equal to the number of layers, let this number be $x$.

Total laddoos = $x \times x = x^2$. (Wait, this is wrong, it's a cubical box with layers)

Let $N$ be the total number of laddoos.

Let $L$ be the number of layers.

Let $P$ be the number of laddoos in each layer.

The problem states $L = P$.

In a cubical arrangement with layers, the number of laddoos in a layer corresponds to the area of the base ($P = \text{width} \times \text{length}$). If it's a cubical box, the width, length, and number of layers are equal. Let this common number be $x$.

Number of layers = $x$ (say, height)

Number of laddoos along length = $x$ (say, length)

Number of laddoos along width = $x$ (say, width)

Number of laddoos in each layer = (Number along length) $\times$ (Number along width) $= x \times x = x^2$.

Total number of laddoos = (Number of laddoos in each layer) $\times$ (Number of layers) $= x^2 \times x = x^3$.

So, the total number of laddoos is $x^3$, where $x$ is the number of layers and also the number of laddoos along one side of a layer.

The case study tells us that the total number of laddoos is 9261, and this number is a perfect cube. The number of layers (and laddoos per side of a layer) is the cube root of 9261.

Total number of laddoos = 9261.

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The question asks for the total number of laddoos.

According to the problem description, this number is explicitly stated as 9261.

Let's verify the calculation mentioned in the case study: $9261$ is indeed a perfect cube. We found in Question 8 that $\sqrt[3]{3375} = 15$. $9261$ is larger. Let's try numbers ending in 1 whose cube ends in 1 (only 1 or 11, 21, etc.). $11^3 = 1331$. $21^3$ ends in the unit digit of $1^3$, which is 1. Let's calculate $21^3$:

$21^2 = 441$

$21^3 = 441 \times 21$

$\begin{array}{cc}& & 4 & 4 & 1 \\ \times & & & 2 & 1 \\ \hline & & 4 & 4 & 1 \\ 8 & 8 & 2 & \times \\ \hline 9 & 2 & 6 & 1 \\ \hline \end{array}$

So, $21^3 = 9261$. The cube root of 9261 is 21. This means there are 21 layers and 21 laddoos along the side of each layer (so $21 \times 21 = 441$ laddoos in each layer). This confirms the structure described in the case study using the number 9261.

The total number of laddoos is simply the number given in the problem, which is 9261.

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The correct option is (A) 9261.

This question explicitly asks to refer to the Case Study presented in Question 20.

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The content of the "Question 20 Case Study" is not included in the provided input.

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To determine the number of layers in the cubical box, information such as the dimensions of the box (particularly its height) and the dimensions or specifications of the items arranged in layers within the box is required.

This necessary information would typically be provided within the text or context of the Case Study mentioned in Question 20.

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Since the details of the Case Study (Question 20) are unavailable, it is not possible to perform any calculations or logical deductions to arrive at the correct number of layers among the given options (19, 21, 23, 27).

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Therefore, a specific numerical answer or derivation cannot be provided without the complete text of the referenced Case Study.

We are asked to find a number whose cube is $13824$. This is equivalent to finding the cube root of $13824$. Let the number be $x$. Then, we need to solve the equation $x^3 = 13824$ for $x$.

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Solution (Using Prime Factorisation):

We can find the cube root of $13824$ by prime factorisation.

Find the prime factors of $13824$:

$\begin{array}{c|cc} 2 & 13824 \\ \hline 2 & 6912 \\ \hline 2 & 3456 \\ \hline 2 & 1728 \\ \hline 2 & 864 \\ \hline 2 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, the prime factorisation of $13824$ is $\$13824 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3\$$.

We can write this as $\$13824 = 2^9 \times 3^3\$$.

To find the cube root, we group the prime factors in triplets:

\$13824 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)\$

\$13824 = 2^3 \times 2^3 \times 2^3 \times 3^3\$

\$13824 = (2 \times 2 \times 2 \times 3)^3\$

\$13824 = (8 \times 3)^3\$

\$13824 = 24^3\$

Therefore, the number whose cube is $13824$ is $24$.

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Alternate Solution (Checking Options):

We can check the cube of each given option to see which one equals $13824$.

(A) $24$:

\$24^3 = 24 \times 24 \times 24\$

\$24 \times 24 = 576\$

\$576 \times 24\$

$\begin{array}{cc}& & 5 & 7 & 6 \\ \times & & & 2 & 4 \\ \hline && 2 & 3 & 0 & 4 \\ & 1 & 1 & 5 & 2 & \times \\ \hline 1 & 3 & 8 & 2 & 4 \\ \hline \end{array}$

So, $\$24^3 = 13824\$$. This matches the given number.

(B) $26$: The last digit of $26$ is $6$. The last digit of $26^3$ is the last digit of $6^3 = 216$, which is $6$. Since $13824$ ends in $4$, $26$ is not the correct answer.

(C) $22$: The last digit of $22$ is $2$. The last digit of $22^3$ is the last digit of $2^3 = 8$. Since $13824$ ends in $4$, $22$ is not the correct answer.

(D) $28$: The last digit of $28$ is $8$. The last digit of $28^3$ is the last digit of $8^3 = 512$, which is $2$. Since $13824$ ends in $4$, $28$ is not the correct answer.

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Both methods confirm that the number whose cube is $13824$ is $24$.

The correct option is (A) $24$.

To find the unit digit of the cube root of a perfect cube, we need to examine the relationship between the unit digit of a number and the unit digit of its cube.

---

Let's list the unit digits of the cubes of the digits from 0 to 9:

$0^3 = 0$ (ends in 0)

$1^3 = 1$ (ends in 1)

$2^3 = 8$ (ends in 8)

$3^3 = 27$ (ends in 7)

$4^3 = 64$ (ends in 4)

$5^3 = 125$ (ends in 5)

$6^3 = 216$ (ends in 6)

$7^3 = 343$ (ends in 3)

$8^3 = 512$ (ends in 2)

$9^3 = 729$ (ends in 9)

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We can observe a pattern here. The unit digit of the cube root of a perfect cube is uniquely determined by the unit digit of the perfect cube itself.

• If a perfect cube ends in 0, its cube root ends in 0.
• If a perfect cube ends in 1, its cube root ends in 1.
• If a perfect cube ends in 8, its cube root ends in 2.
• If a perfect cube ends in 7, its cube root ends in 3.
• If a perfect cube ends in 4, its cube root ends in 4.
• If a perfect cube ends in 5, its cube root ends in 5.
• If a perfect cube ends in 6, its cube root ends in 6.
• If a perfect cube ends in 3, its cube root ends in 7.
• If a perfect cube ends in 2, its cube root ends in 8.
• If a perfect cube ends in 9, its cube root ends in 9.

---

The question states that the perfect cube ends in $8$. Looking at the pattern above, if a perfect cube ends in $8$, its cube root ends in $2$.

---

Therefore, the unit digit of the cube root of a perfect cube ending in 8 is 2.

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Comparing this with the given options:

(A) 2

(B) 8

(C) 4

(D) 6

The correct option is (A).

---

The unit digit of the cube root is 2.

To Find: The number of zeroes at the end of the cube of a number ending with two zeroes.

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Solution:

Let the number ending with two zeroes be $N$.

A number ending with two zeroes can be expressed in the form $k \times 100$, where $k$ is an integer. This can also be written as $k \times 10^2$.

So, let $N = k \times 10^2$.

We need to find the cube of this number, which is $N^3$.

$N^3 = (k \times 10^2)^3$

Using the exponent rule $(ab)^m = a^m b^m$, we get:

$N^3 = k^3 \times (10^2)^3$

Using the exponent rule $(a^m)^n = a^{m \times n}$, we get:

$N^3 = k^3 \times 10^{2 \times 3}$

$N^3 = k^3 \times 10^6$

The term $10^6$ means $1,000,000$. Multiplying $k^3$ by $10^6$ will add 6 zeroes at the end of $k^3$ (assuming $k$ is not 0 or does not make $k^3$ end in zeroes in a way that interferes, but the structure $k \times 10^2$ ensures the trailing zeroes come from the $10^2$ factor). If $k$ is a whole number, $k^3$ will also be a whole number.

For example, let the number be 300. It ends with two zeroes.

$300 = 3 \times 100 = 3 \times 10^2$

The cube is $300^3 = (3 \times 10^2)^3 = 3^3 \times (10^2)^3 = 27 \times 10^6 = 27 \times 1,000,000 = 27,000,000$.

The number $27,000,000$ ends with six zeroes.

Thus, the cube of a number ending with two zeroes will end with $2 \times 3 = 6$ zeroes.

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The number of zeroes is six.

Comparing with the given options, option (D) is correct.

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Final Answer:

The cube of a number ending with two zeroes will end with Six zeroes.

To Find: The step that is NOT valid when finding the cube root of a perfect cube by prime factorization among the given options.

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Solution:

Let's recall the steps for finding the cube root of a perfect cube by prime factorization:

Step 1: Find the prime factorization of the given number.

Step 2: Group the equal prime factors in triplets.

Step 3: From each triplet, take one factor.

Step 4: Multiply the factors taken from each triplet to get the cube root of the number.

Now, let's examine the given options:

(A) Find the prime factors of the number. This corresponds to Step 1, which is a valid step.

(B) Group the prime factors in pairs. This step is used for finding the square root of a number, not the cube root. For cube roots, we group in triplets.

(C) Form triplets of equal prime factors. This corresponds to Step 2, which is a valid step.

(D) Take one factor from each triplet and multiply them. This combines Step 3 and Step 4, which are valid steps.

Therefore, the step that is NOT valid for finding the cube root by prime factorization is grouping the prime factors in pairs.

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The invalid step is (B).

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Final Answer:

The step which is NOT a valid step when finding the cube root of a perfect cube by prime factorization is (B) Group the prime factors in pairs.

To Find: The number of digits in the cube root of 753571.

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Solution:

We can determine the number of digits in the cube root of a perfect cube by grouping the digits of the number from the right in groups of three.

The given number is 753571.

We group the digits from the right in triplets:

$\overline{753} \; \overline{571}$

There are two groups of digits.

The number of digits in the cube root is equal to the number of groups formed.

Since there are 2 groups, the cube root of 753571 will have 2 digits.

Alternatively, we can consider the number of digits in the original number.

The number 753571 has 6 digits.

We know that:

The cube root of a number with 1, 2, or 3 digits has 1 digit.

The cube root of a number with 4, 5, or 6 digits has 2 digits.

The cube root of a number with 7, 8, or 9 digits has 3 digits.

Since 753571 has 6 digits, its cube root will have 2 digits.

For verification, let's consider the bounds for 2-digit cube roots:

$10^3 = 1000$

$99^3 = 970299$

Since $1000 \leq 753571 < 970299$, the cube root $\sqrt[3]{753571}$ must be a 2-digit number between 10 and 99.

In fact, $91^3 = 753571$, which is a 2-digit number.

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The number of digits in the cube root is 2.

Comparing with the given options, option (B) is correct.

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Final Answer:

The cube root of 753571 will have 2 digits.

To Find: The cube root of $0.125$.

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Solution:

We need to find the value of $\sqrt[3]{0.125}$.

First, let's convert the decimal number $0.125$ into a fraction.

$0.125 = \frac{125}{1000}$

Now, we need to find the cube root of this fraction:

$\sqrt[3]{0.125} = \sqrt[3]{\frac{125}{1000}}$

Using the property of cube roots, $\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}$, we can write:

$\sqrt[3]{\frac{125}{1000}} = \frac{\sqrt[3]{125}}{\sqrt[3]{1000}}$

Now, we find the cube root of the numerator (125) and the denominator (1000).

To find the cube root of 125, we find its prime factors:

$125 = 5 \times 5 \times 5 = 5^3$

So, $\sqrt[3]{125} = 5$.

To find the cube root of 1000, we find its prime factors:

$1000 = 10 \times 10 \times 10 = 10^3$

So, $\sqrt[3]{1000} = 10$.

Substitute these values back into the fraction:

$\frac{\sqrt[3]{125}}{\sqrt[3]{1000}} = \frac{5}{10}$

Now, simplify the fraction $\frac{5}{10}$ and convert it back to a decimal.

$\frac{5}{10} = \frac{1}{2} = 0.5$

Thus, the cube root of $0.125$ is $0.5$.

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Comparing the result with the given options:

(A) 0.5 - Matches our result.

(B) 0.05 - Incorrect ($0.05^3 = 0.000125$).

(C) 0.25 - Incorrect ($0.25^3 = 0.015625$).

(D) 0.1 - Incorrect ($0.1^3 = 0.001$).

The correct option is (A).

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Final Answer:

The cube root of $0.125$ is 0.5.

To Find: The side length of the new cube formed by melting and reforming three cubes with given side lengths.

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Solution:

When the three cubes are melted and reformed into a single cube, the total volume of the material remains constant.

Let the side lengths of the three cubes be $s_1 = 3 \text{ cm}$, $s_2 = 4 \text{ cm}$, and $s_3 = 5 \text{ cm}$.

The volume of a cube with side length $s$ is given by the formula $V = s^3$.

Volume of the first cube, $V_1 = s_1^3 = (3 \text{ cm})^3 = 3^3 \text{ cm}^3 = 27 \text{ cm}^3$.

Volume of the second cube, $V_2 = s_2^3 = (4 \text{ cm})^3 = 4^3 \text{ cm}^3 = 64 \text{ cm}^3$.

Volume of the third cube, $V_3 = s_3^3 = (5 \text{ cm})^3 = 5^3 \text{ cm}^3 = 125 \text{ cm}^3$.

The total volume of the three cubes is the sum of their individual volumes:

$V_{\text{total}} = V_1 + V_2 + V_3$

$V_{\text{total}} = 27 \text{ cm}^3 + 64 \text{ cm}^3 + 125 \text{ cm}^3$

$V_{\text{total}} = (27 + 64 + 125) \text{ cm}^3$

$V_{\text{total}} = (91 + 125) \text{ cm}^3$

$V_{\text{total}} = 216 \text{ cm}^3$.

Let the side length of the new single cube be $S \text{ cm}$. The volume of the new cube is $V_{\text{new}} = S^3$.

Since the total volume is conserved, $V_{\text{new}} = V_{\text{total}}$.

$S^3 = 216 \text{ cm}^3$

To find the side length $S$, we need to calculate the cube root of 216.

$S = \sqrt[3]{216}$

We find the prime factorization of 216:

$\begin{array}{c|cc} 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^3 \times 3^3 = (2 \times 3)^3 = 6^3$.

So, $S = \sqrt[3]{6^3} = 6$.

The side length of the new cube is 6 cm.

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Comparing the result with the given options:

(A) $6 \text{ cm}$ - Matches our result.

(B) $7 \text{ cm}$ - Incorrect ($7^3 = 343$).

(C) $12 \text{ cm}$ - Incorrect ($12^3 = 1728$).

(D) $15 \text{ cm}$ - Incorrect ($15^3 = 3375$).

The correct option is (A).

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Final Answer:

The side length of the new cube is $6 \text{ cm}$.

To Find: The property which is FALSE regarding the cube of a number among the given options.

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Solution:

Let's examine each property:

(A) The cube of a number is the number multiplied by itself three times.

This is the definition of a cube. For any number $x$, its cube is $x \times x \times x = x^3$. This property is TRUE.

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(B) The sum of cubes of first $n$ natural numbers is equal to the square of the sum of first $n$ natural numbers, i.e., $1^3 + 2^3 + ... + n^3 = (1+2+...+n)^2$.

This is a well-known mathematical identity. The sum of the first $n$ natural numbers is given by $\frac{n(n+1)}{2}$. The identity states that $\sum_{k=1}^{n} k^3 = \left(\sum_{k=1}^{n} k\right)^2$. For example, for $n=2$: $1^3 + 2^3 = 1 + 8 = 9$. $(1+2)^2 = 3^2 = 9$. For $n=3$: $1^3 + 2^3 + 3^3 = 1 + 8 + 27 = 36$. $(1+2+3)^2 = 6^2 = 36$. This property is TRUE.

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(C) There is no perfect cube ending in 3 or 7.

Let's consider the last digit of cubes of numbers ending in different digits:

• Numbers ending in 1, their cube ends in 1 ($1^3=1, 11^3=1331$).
• Numbers ending in 2, their cube ends in 8 ($2^3=8, 12^3=1728$).
• Numbers ending in 3, their cube ends in 7 ($3^3=27, 13^3=2197$).
• Numbers ending in 4, their cube ends in 4 ($4^3=64, 14^3=2744$).
• Numbers ending in 5, their cube ends in 5 ($5^3=125, 15^3=3375$).
• Numbers ending in 6, their cube ends in 6 ($6^3=216, 16^3=4096$).
• Numbers ending in 7, their cube ends in 3 ($7^3=343, 17^3=4913$).
• Numbers ending in 8, their cube ends in 2 ($8^3=512, 18^3=5832$).
• Numbers ending in 9, their cube ends in 9 ($9^3=729, 19^3=6859$).
• Numbers ending in 0, their cube ends in 0 ($10^3=1000, 20^3=8000$).

As seen from the examples, perfect cubes can end in 3 (if the original number ends in 7) and perfect cubes can end in 7 (if the original number ends in 3). Therefore, the statement "There is no perfect cube ending in 3 or 7" is FALSE.

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(D) The cube of a multiple of 3 is a multiple of 27.

Let the number be a multiple of 3, so it can be written as $3k$ for some integer $k$. The cube of this number is $(3k)^3$.

$(3k)^3 = 3^3 \times k^3 = 27 \times k^3$

Since $k^3$ is an integer, $27 \times k^3$ is clearly a multiple of 27. For example, the cube of 6 (which is a multiple of 3) is $6^3 = 216$. $216 = 27 \times 8$, so 216 is a multiple of 27. This property is TRUE.

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Based on the analysis, the FALSE property is (C).

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Final Answer:

The property which is FALSE regarding the cube of a number is (C) There is no perfect cube ending in 3 or 7.

To Find: The volume of the tank based on the information provided in the case study.

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Solution:

The case study explicitly states the volume of the tank.

The sentence "A large cubical storage tank in Chennai has a volume of $216 \text{ m}^3$" directly gives the volume.

Therefore, the volume of the tank is $216 \text{ m}^3$.

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Comparing the given options:

(A) $216 \text{ m}^3$ - This matches the volume stated in the case study.

(B) $\sqrt{216} \text{ m}^3$ - This represents the square root of the volume, not the volume itself.

(C) $\sqrt[3]{216} \text{ m}^3$ - This represents the cube root of the volume, which is the side length ($6 \text{ m}$), not the volume itself.

(D) $216^3 \text{ m}^3$ - This represents the cube of the volume, which is much larger than the given volume.

The correct option is (A).

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Final Answer:

The volume of the tank is $216 \text{ m}^3$.

To Find: The side length of the cubical tank.

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Given:

Volume of the cubical tank = $216 \text{ m}^3$.

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Solution:

For a cubical tank with side length $s$, the volume $V$ is given by the formula $V = s^3$.

We are given that the volume is $216 \text{ m}^3$. So, we have:

$s^3 = 216 \text{ m}^3$

To find the side length $s$, we need to find the cube root of the volume.

$s = \sqrt[3]{216}$

We need to find a number whose cube is 216.

We can find the cube root by prime factorization of 216:

$\begin{array}{c|cc} 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The prime factorization of 216 is $2 \times 2 \times 2 \times 3 \times 3 \times 3$.

We group the prime factors in triplets:

$216 = (2 \times 2 \times 2) \times (3 \times 3 \times 3) = 2^3 \times 3^3$

Now, we take the cube root:

$s = \sqrt[3]{2^3 \times 3^3} = \sqrt[3]{(2 \times 3)^3} = \sqrt[3]{6^3}$

$s = 6$

The side length of the tank is 6 meters.

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Comparing the result with the given options:

(A) $6 \text{ m}$ - Matches our result.

(B) $8 \text{ m}$ - Incorrect ($8^3 = 512$).

(C) $12 \text{ m}$ - Incorrect ($12^3 = 1728$).

(D) $16 \text{ m}$ - Incorrect ($16^3 = 4096$).

The correct option is (A).

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Final Answer:

The side length of the cubical tank is $6 \text{ m}$.

We are asked to evaluate the expression $\sqrt[3]{-8} + \sqrt[3]{27}$.

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First, let's evaluate the cube root of -8. The cube root of a number is the value that, when multiplied by itself three times, gives the original number.

We are looking for a number $x$ such that $x^3 = -8$.

Let's test a negative integer. Consider $-2$.

$(-2) \times (-2) \times (-2) = (4) \times (-2) = -8$.

So, the cube root of -8 is -2.

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Next, let's evaluate the cube root of 27.

We are looking for a number $y$ such that $y^3 = 27$.

Let's test positive integers. Consider 3.

$3 \times 3 \times 3 = 9 \times 3 = 27$.

So, the cube root of 27 is 3.

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Now, substitute these values back into the original expression:

$\sqrt[3]{-8} + \sqrt[3]{27} = -2 + 3$

Performing the addition:

$-2 + 3 = 1$

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Comparing our result with the given options:

(A) $-2 + 3 = 1$

(B) $-2 - 3 = -5$

(C) $2 + 3 = 5$

(D) $-4 + 9 = 5$

Our calculated value, 1, matches the result in option (A).

---

Therefore, the correct statement is $-2 + 3 = 1$.

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The correct option is (A).

To Find: The value of $x$ given the equation $x^3 = 729$.

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Solution:

We are given the equation $x^3 = 729$.

To find the value of $x$, we need to calculate the cube root of 729.

$x = \sqrt[3]{729}$

We can find the cube root by prime factorization of 729:

$\begin{array}{c|cc} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The prime factorization of 729 is $3 \times 3 \times 3 \times 3 \times 3 \times 3$.

We group the prime factors in triplets:

$729 = (3 \times 3 \times 3) \times (3 \times 3 \times 3) = 3^3 \times 3^3$

Using the exponent rule $a^m \times a^n = a^{m+n}$, we get $3^3 \times 3^3 = 3^{3+3} = 3^6$.

So, $729 = 3^6$.

To find the cube root, we can write $729$ as a cube:

$729 = (3^2)^3 = 9^3$

So, the equation $x^3 = 729$ becomes $x^3 = 9^3$.

Taking the cube root of both sides:

$x = \sqrt[3]{9^3}$

$x = 9$

Alternatively, using the prime factorization $729 = 3^6$:

$x = \sqrt[3]{3^6} = 3^{6/3} = 3^2 = 9$.

Thus, the value of $x$ is 9.

---

Comparing the result with the given options:

(A) 9 - Matches our result.

(B) 19 - Incorrect ($19^3 = 6859$).

(C) 29 - Incorrect ($29^3 = 24389$).

(D) 81 - Incorrect ($81^3$ is much larger). Note that $81 = 9^2$, not $9^3$ or $\sqrt[3]{729}$.

The correct option is (A).

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Final Answer:

The value of $x$ is 9.

We need to evaluate the truthfulness of the given statements about cube roots.

---

Let's analyze each statement:

(A) The cube root of a negative number is negative.

Consider a negative number, say -8. We look for a number $x$ such that $x^3 = -8$. We know that $(-2)^3 = (-2) \times (-2) \times (-2) = 4 \times (-2) = -8$. Thus, $\sqrt[3]{-8} = -2$, which is a negative number. In general, for any negative number $a$, its cube root $\sqrt[3]{a}$ is real and negative because the cube of a positive number is positive, and the cube of zero is zero.

This statement is TRUE.

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(B) The cube root of a number is the number raised to the power of $\frac{1}{3}$.

By the definition of rational exponents, for any real number $a$ and a positive integer $n$, $a^{1/n}$ is defined as the $n$-th root of $a$, provided the $n$-th root exists. For $n=3$, $a^{1/3}$ is the cube root of $a$.

This statement is TRUE.

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(C) The cube root of a perfect cube is always an integer.

A perfect cube is a number that is the cube of an integer. For example, $8 = 2^3$, $27 = 3^3$, $-64 = (-4)^3$, $0 = 0^3$. If a number is a perfect cube, it can be written as $k^3$ for some integer $k$. The cube root of $k^3$ is $\sqrt[3]{k^3}$. For real numbers, $\sqrt[3]{k^3} = k$. Since $k$ is an integer, $\sqrt[3]{k^3}$ is also an integer.

This statement is TRUE.

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(D) The symbol for cube root is $\sqrt[3]{}$.

The symbol $\sqrt{}$ is the radical sign, used for roots. The index of the root is placed above the radical sign. For the cube root, the index is 3. Therefore, the symbol $\sqrt[3]{}$ represents the cube root.

This is the standard mathematical notation for the cube root.

This statement is TRUE.

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Since all four statements are true, we select all of them.

The correct options are (A), (B), (C), and (D).

We are given an Assertion (A) and a Reason (R) and need to determine their truthfulness and the relationship between them.

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Assertion (A): The cube root of 1000 is 10.

The cube root of a number $x$, denoted by $\sqrt[3]{x}$, is the number $y$ such that $y^3 = x$.

To check if the cube root of 1000 is 10, we need to verify if $10^3$ equals 1000.

$10^3 = 10 \times 10 \times 10 = 100 \times 10 = 1000$.

Since $10^3 = 1000$, it is true that the cube root of 1000 is 10.

Thus, Assertion (A) is TRUE.

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Reason (R): $10^3 = 1000$.

We have already calculated $10^3$ in the evaluation of Assertion (A):

$10^3 = 10 \times 10 \times 10 = 1000$.

Thus, Reason (R) is also TRUE.

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Now, we need to determine if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states that $\sqrt[3]{1000} = 10$. The definition of the cube root tells us that $\sqrt[3]{x} = y$ if and only if $y^3 = x$.

In this case, $x=1000$ and $y=10$. Assertion (A) claims $\sqrt[3]{1000} = 10$. According to the definition, this is true if $10^3 = 1000$.

Reason (R) states exactly that: $10^3 = 1000$.

Therefore, Reason (R) provides the fundamental mathematical relationship that confirms the truth of Assertion (A). R directly explains why 10 is the cube root of 1000.

---

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A).

---

This matches option (A).

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The correct option is (A).

The question asks to complete the sentence: "The number whose cube is $x$ is called the _________ of $x$."

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Let the number whose cube is $x$ be $y$.

According to the statement, the cube of this number $y$ is equal to $x$. Mathematically, this can be written as:

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By the definition of a cube root, if $y^3 = x$, then $y$ is called the cube root of $x$. This is denoted by $y = \sqrt[3]{x}$ or $y = x^{1/3}$.

---

Therefore, the number whose cube is $x$ is called the cube root of $x$.

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Let's look at the options provided:

(A) Square: The number whose square is $x$ is called the square root of $x$. ($y^2 = x \implies y = \sqrt{x}$)

(B) Square root: This is the answer to the number whose square is $x$.

(C) Cube: The cube of a number $y$ is $y^3$. $x$ is the cube, and $y$ is the number.

(D) Cube root: This matches our definition derived from the statement $y^3 = x$.

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The correct term to complete the sentence is "Cube root".

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The correct option is (D).

To determine the unit digit of a perfect cube, we only need to look at the unit digit of the number being cubed. Let's examine the unit digits of the cubes of the digits from 0 to 9.

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Consider a number ending in a specific digit. Its cube will have a unit digit determined by the cube of that unit digit.

Unit digit of Number	Cube	Unit digit of Cube
0	$0^3 = 0$	0
1	$1^3 = 1$	1
2	$2^3 = 8$	8
3	$3^3 = 27$	7
4	$4^3 = 64$	4
5	$5^3 = 125$	5
6	$6^3 = 216$	6
7	$7^3 = 343$	3
8	$8^3 = 512$	2
9	$9^3 = 729$	9

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The possible unit digits of a perfect cube are the unit digits found in the third column of the table: 0, 1, 8, 7, 4, 5, 6, 3, 2, 9.

This set of possible unit digits is $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. It includes all the digits from 0 to 9.

---

Now let's look at the given options:

(A) 3: From the table, the cube of a number ending in 7 has a unit digit of 3. For example, $7^3=343$. So, 3 can be the unit digit of a perfect cube.

(B) 4: From the table, the cube of a number ending in 4 has a unit digit of 4. For example, $4^3=64$. So, 4 can be the unit digit of a perfect cube.

(C) 5: From the table, the cube of a number ending in 5 has a unit digit of 5. For example, $5^3=125$. So, 5 can be the unit digit of a perfect cube.

(D) 6: From the table, the cube of a number ending in 6 has a unit digit of 6. For example, $6^3=216$. So, 6 can be the unit digit of a perfect cube.

---

Based on the mathematical properties of perfect cubes, any digit from 0 to 9 can be the unit digit of a perfect cube. Therefore, all the options provided (3, 4, 5, and 6) can be unit digits of perfect cubes.

---

As the question asks which of the following cannot be the unit digit, and our analysis shows that all options can be the unit digit, it appears there might be an error in the question or the provided options, as none of the given options satisfy the condition of not being a unit digit of a perfect cube based on standard mathematical properties.

We are asked to find the cube of the fraction $\frac{-1}{4}$.

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The cube of a number is the result of multiplying the number by itself three times. For a fraction $\frac{p}{q}$, its cube is $\left(\frac{p}{q}\right)^3$, which is equal to $\frac{p^3}{q^3}$.

---

In this problem, the number is $\frac{-1}{4}$. So we need to calculate $\left(\frac{-1}{4}\right)^3$.

Using the rule for cubing a fraction, we cube the numerator and the denominator separately:

---

Now, let's calculate the cube of the numerator, which is $-1$:

Since $(-1) \times (-1) = 1$, we have:

---

Next, let's calculate the cube of the denominator, which is $4$:

Since $4 \times 4 = 16$, we have:

---

Now, substitute the results back into the fraction:

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Comparing our result with the given options:

(A) $\frac{1}{64}$

(B) $-\frac{1}{12}$

(C) $-\frac{1}{64}$

(D) $\frac{1}{12}$

Our calculated value, $-\frac{1}{64}$, matches option (C).

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The correct option is (C).

A cube number or a perfect cube is a natural number that is the cube of some natural number. In other words, a number $n$ is a perfect cube if there exists a natural number $m$ such that $n = m \times m \times m = m^3$.

---

Here are four examples of perfect cubes:

The cube of 1 is $1^3 = 1 \times 1 \times 1 = 1$. So, 1 is a perfect cube.

The cube of 2 is $2^3 = 2 \times 2 \times 2 = 8$. So, 8 is a perfect cube.

The cube of 3 is $3^3 = 3 \times 3 \times 3 = 27$. So, 27 is a perfect cube.

The cube of 4 is $4^3 = 4 \times 4 \times 4 = 64$. So, 64 is a perfect cube.

(a) $216$

To check if 216 is a perfect cube, we find its prime factorization:

$\begin{array}{c|cc} 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^3 \times 3^3$.

Since the exponents of all prime factors (2 and 3) are multiples of 3, 216 is a perfect cube. Specifically, $216 = (2 \times 3)^3 = 6^3$.

---

(b) $300$

To check if 300 is a perfect cube, we find its prime factorization:

$\begin{array}{c|cc} 2 & 300 \\ \hline 2 & 150 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, $300 = 2 \times 2 \times 3 \times 5 \times 5 = 2^2 \times 3^1 \times 5^2$.

Since the exponents of the prime factors (2, 3, and 5) are 2, 1, and 2, which are not multiples of 3, 300 is not a perfect cube.

---

(c) $1000$

To check if 1000 is a perfect cube, we find its prime factorization:

$\begin{array}{c|cc} 2 & 1000 \\ \hline 2 & 500 \\ \hline 2 & 250 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, $1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 = 2^3 \times 5^3$.

Since the exponents of all prime factors (2 and 5) are multiples of 3, 1000 is a perfect cube. Specifically, $1000 = (2 \times 5)^3 = 10^3$.

---

(d) $400$

To check if 400 is a perfect cube, we find its prime factorization:

$\begin{array}{c|cc} 2 & 400 \\ \hline 2 & 200 \\ \hline 2 & 100 \\ \hline 2 & 50 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, $400 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 = 2^4 \times 5^2$.

Since the exponents of the prime factors (2 and 5) are 4 and 2, which are not multiples of 3, 400 is not a perfect cube.

Here are three properties of the cubes of natural numbers relating to their unit digit:

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Property 1: If a natural number ends with the digit 0, 1, 4, 5, 6, or 9, then its cube also ends with the same digit.

For example:

$1^3 = 1$ (ends in 1)

$4^3 = 64$ (ends in 4)

$5^3 = 125$ (ends in 5)

$6^3 = 216$ (ends in 6)

$9^3 = 729$ (ends in 9)

$10^3 = 1000$ (ends in 0)

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Property 2: If a natural number ends with the digit 2, then its cube ends with the digit 8.

For example:

$2^3 = 8$ (ends in 8)

$12^3 = 1728$ (ends in 8)

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Property 3: If a natural number ends with the digit 3, then its cube ends with the digit 7.

For example:

$3^3 = 27$ (ends in 7)

$13^3 = 2197$ (ends in 7)

(a) $32$

The unit digit of 32 is 2.

The unit digit of the cube of a number is the same as the unit digit of the cube of its unit digit.

The cube of the unit digit 2 is $2^3 = 8$.

Therefore, the unit digit of the cube of 32 is 8.

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(b) $59$

The unit digit of 59 is 9.

The unit digit of the cube of a number is the same as the unit digit of the cube of its unit digit.

The cube of the unit digit 9 is $9^3 = 729$. The unit digit of 729 is 9.

Therefore, the unit digit of the cube of 59 is 9.

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(c) $101$

The unit digit of 101 is 1.

The unit digit of the cube of a number is the same as the unit digit of the cube of its unit digit.

The cube of the unit digit 1 is $1^3 = 1$. The unit digit of 1 is 1.

Therefore, the unit digit of the cube of 101 is 1.

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(d) $408$

The unit digit of 408 is 8.

The unit digit of the cube of a number is the same as the unit digit of the cube of its unit digit.

The cube of the unit digit 8 is $8^3 = 512$. The unit digit of 512 is 2.

Therefore, the unit digit of the cube of 408 is 2.

To find the number of perfect cubes between 1 and 100, we need to find the natural numbers whose cubes are greater than 1 and less than 100.

---

Let's list the cubes of the first few natural numbers:

$1^3 = 1$ (Not between 1 and 100, as it is equal to 1)

$2^3 = 8$ (Between 1 and 100)

$3^3 = 27$ (Between 1 and 100)

$4^3 = 64$ (Between 1 and 100)

$5^3 = 125$ (Not between 1 and 100, as it is greater than 100)

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Since $5^3$ is already greater than 100, any subsequent cube $(6^3, 7^3, ...)$ will also be greater than 100.

The perfect cubes between 1 and 100 are 8, 27, and 64.

There are 3 perfect cubes between 1 and 100.

Yes, the cube of an odd number is always odd.

Justification:

An odd number can be represented in the form $2k+1$ for some integer $k$.

The cube of an odd number is $(2k+1)^3$.

Using the formula $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$, we have:

$(2k+1)^3 = (2k)^3 + 3(2k)^2(1) + 3(2k)(1)^2 + 1^3$

$(2k+1)^3 = 8k^3 + 3(4k^2)(1) + 3(2k)(1) + 1$

$(2k+1)^3 = 8k^3 + 12k^2 + 6k + 1$

We can factor out 2 from the first three terms:

$(2k+1)^3 = 2(4k^3 + 6k^2 + 3k) + 1$

Since $4k^3 + 6k^2 + 3k$ is an integer (let's call it $m$), the cube is in the form $2m+1$, which is the definition of an odd number.

Examples:

$1^3 = 1$ (odd)

$3^3 = 27$ (odd)

$5^3 = 125$ (odd)

$7^3 = 343$ (odd)

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Yes, the cube of an even number is always even.

Justification:

An even number can be represented in the form $2k$ for some integer $k$.

The cube of an even number is $(2k)^3$.

$(2k)^3 = 2^3 \times k^3 = 8k^3$

We can write $8k^3$ as $2(4k^3)$.

Since $4k^3$ is an integer (let's call it $m$), the cube is in the form $2m$, which is the definition of an even number.

Examples:

$2^3 = 8$ (even)

$4^3 = 64$ (even)

$6^3 = 216$ (even)

$8^3 = 512$ (even)

(a) $-4$

The cube of $-4$ is calculated as $(-4) \times (-4) \times (-4)$.

$(-4)^3 = (-4) \times (-4) \times (-4)$

First, multiply the first two terms: $(-4) \times (-4) = 16$ (since negative times negative is positive).

Next, multiply the result by the third term: $16 \times (-4) = -64$ (since positive times negative is negative).

Therefore, the cube of $-4$ is $-64$.

$(-4)^3 = -64$

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(b) $-7$

The cube of $-7$ is calculated as $(-7) \times (-7) \times (-7)$.

$(-7)^3 = (-7) \times (-7) \times (-7)$

First, multiply the first two terms: $(-7) \times (-7) = 49$ (since negative times negative is positive).

Next, multiply the result by the third term: $49 \times (-7) = -343$ (since positive times negative is negative).

Therefore, the cube of $-7$ is $-343$.

$(-7)^3 = -343$

(a) $\frac{2}{5}$

The cube of $\frac{2}{5}$ is $\left(\frac{2}{5}\right)^3$.

To find the cube of a fraction, we cube the numerator and the denominator separately.

$\left(\frac{2}{5}\right)^3 = \frac{2^3}{5^3}$

$2^3 = 2 \times 2 \times 2 = 8$

$5^3 = 5 \times 5 \times 5 = 125$

So, $\left(\frac{2}{5}\right)^3 = \frac{8}{125}$.

The cube of $\frac{2}{5}$ is $\frac{8}{125}$.

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(b) $\frac{-1}{3}$

The cube of $\frac{-1}{3}$ is $\left(\frac{-1}{3}\right)^3$.

To find the cube of a fraction, we cube the numerator and the denominator separately.

$\left(\frac{-1}{3}\right)^3 = \frac{(-1)^3}{3^3}$

$(-1)^3 = (-1) \times (-1) \times (-1) = 1 \times (-1) = -1$

$3^3 = 3 \times 3 \times 3 = 27$

So, $\left(\frac{-1}{3}\right)^3 = \frac{-1}{27}$.

The cube of $\frac{-1}{3}$ is $\frac{-1}{27}$.

The cube root of a number $x$ is the number $y$ such that when $y$ is cubed (multiplied by itself three times), the result is $x$.

In mathematical terms, if $y^3 = x$, then $y$ is the cube root of $x$.

For example, the cube root of 8 is 2 because $2^3 = 2 \times 2 \times 2 = 8$.

---

The symbol used for the cube root is $\sqrt[3]{\phantom{x}}$. The small digit '3' above the square root symbol indicates that it is a cube root.

So, the cube root of $x$ is written as $\sqrt[3]{x}$.

To find the cube root of 64 by the prime factorization method, we first find the prime factors of 64.

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Prime factorization of 64:

$\begin{array}{c|cc} 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

So, the prime factorization of 64 is $2 \times 2 \times 2 \times 2 \times 2 \times 2$.

$64 = 2^6$

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To find the cube root, we group the prime factors into triplets of equal factors.

$64 = (2 \times 2 \times 2) \times (2 \times 2 \times 2)$

$64 = 2^3 \times 2^3$

---

Now, we take one factor from each group and multiply them together.

$\sqrt[3]{64} = \sqrt[3]{2^3 \times 2^3}$

$\sqrt[3]{64} = 2 \times 2$

$\sqrt[3]{64} = 4$

---

Therefore, the cube root of 64 is 4.

To find the cube root of 343 by the prime factorization method, we first find the prime factors of 343.

---

Prime factorization of 343:

We test small prime numbers. 343 is not divisible by 2 (it's odd). The sum of digits is $3+4+3=10$, which is not divisible by 3. It doesn't end in 0 or 5, so not divisible by 5. Let's try 7.

$343 \div 7$

$343 = 7 \times 49$

$49 = 7 \times 7$

So, the prime factors of 343 are 7, 7, and 7.

$\begin{array}{c|cc} 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

So, the prime factorization of 343 is $7 \times 7 \times 7$.

$343 = 7^3$

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To find the cube root, we group the prime factors into triplets of equal factors.

$343 = (7 \times 7 \times 7)$

$343 = 7^3$

---

Now, we take one factor from the group.

$\sqrt[3]{343} = \sqrt[3]{7^3}$

$\sqrt[3]{343} = 7$

---

Therefore, the cube root of 343 is 7.

To find the cube root of a negative number, we find the cube root of the corresponding positive number and then put a negative sign in front of the result.

So, we need to find $\sqrt[3]{-216} = - \sqrt[3]{216}$.

---

First, we find the prime factorization of 216.

Prime factorization of 216:

$\begin{array}{c|cc} 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, the prime factorization of 216 is $2 \times 2 \times 2 \times 3 \times 3 \times 3$.

$216 = 2^3 \times 3^3$

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To find the cube root of 216, we group the prime factors into triplets of equal factors.

$216 = (2 \times 2 \times 2) \times (3 \times 3 \times 3)$

$216 = 2^3 \times 3^3$

---

Now, we take one factor from each group and multiply them together to find $\sqrt[3]{216}$.

$\sqrt[3]{216} = \sqrt[3]{2^3 \times 3^3}$

$\sqrt[3]{216} = 2 \times 3$

$\sqrt[3]{216} = 6$

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Finally, we apply the negative sign for the cube root of -216.

$\sqrt[3]{-216} = - \sqrt[3]{216} = -6$

Therefore, the cube root of -216 is -6.

To find the smallest number by which 108 must be multiplied to get a perfect cube, we first find the prime factorization of 108.

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Prime factorization of 108:

$\begin{array}{c|cc} 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, the prime factorization of 108 is $2 \times 2 \times 3 \times 3 \times 3$.

$108 = 2^2 \times 3^3$

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To make 108 a perfect cube, the exponents of all its prime factors must be multiples of 3.

In the factorization $2^2 \times 3^3$, the prime factor 3 already has an exponent that is a multiple of 3 (which is 3 itself). However, the prime factor 2 has an exponent of 2, which is not a multiple of 3.

To make the exponent of 2 a multiple of 3 (the smallest being 3), we need one more factor of 2 ($2^1$).

$2^2 \times 2^1 = 2^{2+1} = 2^3$

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So, we need to multiply 108 by 2 to make it a perfect cube.

New number = $108 \times 2 = (2^2 \times 3^3) \times 2^1 = 2^3 \times 3^3 = (2 \times 3)^3 = 6^3 = 216$.

216 is a perfect cube ($6^3$).

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The smallest number by which 108 must be multiplied to get a perfect cube is 2.

To find the smallest number by which 81 must be divided to get a perfect cube, we first find the prime factorization of 81.

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Prime factorization of 81:

$\begin{array}{c|cc} 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, the prime factorization of 81 is $3 \times 3 \times 3 \times 3$.

$81 = 3^4$

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For a number to be a perfect cube, the exponents of all its prime factors in the prime factorization must be multiples of 3.

In the factorization $3^4$, the exponent of the prime factor 3 is 4, which is not a multiple of 3.

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To make the exponent of 3 a multiple of 3 by dividing, we need to remove the excess factors of 3.

The exponent is 4. The largest multiple of 3 less than 4 is 3. So, we need to reduce the exponent from 4 to 3.

This means we need to divide by $3^{4-3} = 3^1 = 3$.

---

If we divide 81 by 3, we get $81 \div 3 = 27$.

The prime factorization of 27 is $3 \times 3 \times 3 = 3^3$. This is a perfect cube.

---

Therefore, the smallest number by which 81 must be divided to get a perfect cube is 3.

We need to find the value of $\sqrt[3]{8 \times 125}$.

---

We can use the property of cube roots which states that $\sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b}$.

Applying this property, we get:

$\sqrt[3]{8 \times 125} = \sqrt[3]{8} \times \sqrt[3]{125}$

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Now, we find the cube root of 8 and the cube root of 125 separately.

The prime factorization of 8 is $2 \times 2 \times 2 = 2^3$.

So, $\sqrt[3]{8} = \sqrt[3]{2^3} = 2$.

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The prime factorization of 125 is $5 \times 5 \times 5 = 5^3$.

So, $\sqrt[3]{125} = \sqrt[3]{5^3} = 5$.

---

Now, we multiply the cube roots we found:

$\sqrt[3]{8} \times \sqrt[3]{125} = 2 \times 5 = 10$

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Alternatively, we could first calculate the product $8 \times 125$.

$8 \times 125 = 1000$

Then, we find the cube root of 1000.

The prime factorization of 1000 is $10 \times 10 \times 10 = 10^3$.

So, $\sqrt[3]{1000} = \sqrt[3]{10^3} = 10$.

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In both methods, the result is the same.

Therefore, the cube root of $8 \times 125$ is 10.

We need to find the value of $\sqrt[3]{\frac{27}{64}}$.

---

We can use the property of cube roots which states that $\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}$.

Applying this property, we get:

$\sqrt[3]{\frac{27}{64}} = \frac{\sqrt[3]{27}}{\sqrt[3]{64}}$

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Now, we find the cube root of the numerator (27) and the cube root of the denominator (64) separately.

First, find $\sqrt[3]{27}$. We find the prime factorization of 27:

$\begin{array}{c|cc} 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $27 = 3 \times 3 \times 3 = 3^3$.

Therefore, $\sqrt[3]{27} = \sqrt[3]{3^3} = 3$.

---

Next, find $\sqrt[3]{64}$. We find the prime factorization of 64:

$\begin{array}{c|cc} 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

So, $64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6$.

We can group the factors in triplets: $64 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) = 2^3 \times 2^3 = (2 \times 2)^3 = 4^3$.

Therefore, $\sqrt[3]{64} = \sqrt[3]{4^3} = 4$.

---

Now, substitute the cube roots back into the fraction:

$\frac{\sqrt[3]{27}}{\sqrt[3]{64}} = \frac{3}{4}$

---

Therefore, the cube root of $\frac{27}{64}$ is $\frac{3}{4}$.

We need to find the value of $\sqrt[3]{-1000}$.

---

The cube root of a negative number is equal to the negative of the cube root of the corresponding positive number.

So, $\sqrt[3]{-1000} = -\sqrt[3]{1000}$.

---

Now, we find the cube root of 1000.

We find the prime factorization of 1000:

$\begin{array}{c|cc} 2 & 1000 \\ \hline 2 & 500 \\ \hline 2 & 250 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, the prime factorization of 1000 is $2 \times 2 \times 2 \times 5 \times 5 \times 5$.

$1000 = 2^3 \times 5^3$

---

To find the cube root, we group the prime factors into triplets of equal factors.

$1000 = (2 \times 2 \times 2) \times (5 \times 5 \times 5)$

$1000 = 2^3 \times 5^3$

$\sqrt[3]{1000} = \sqrt[3]{2^3 \times 5^3}$

$\sqrt[3]{1000} = 2 \times 5$

$\sqrt[3]{1000} = 10$

---

Finally, we apply the negative sign to the cube root of 1000 to find the cube root of -1000.

$\sqrt[3]{-1000} = - \sqrt[3]{1000} = -10$

---

Therefore, the cube root of -1000 is -10.

To estimate the cube root of 4913, we follow these steps:

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Step 1: Group the digits of the number from the right in sets of three.

Starting from the rightmost digit, we form groups of three digits. The number is 4913.

The first group from the right is $\mathbf{913}$.

The remaining digits form the second group, which is $\mathbf{4}$.

The groups are 4 and 913.

---

Step 2: Determine the unit digit of the cube root.

Consider the first group from the right, which is 913.

The unit digit of 913 is 3.

Now, we look at the unit digits of the cubes of natural numbers from 1 to 10:

$1^3 = 1$

$2^3 = 8$

$3^3 = 27$ (unit digit is 7)

$4^3 = 64$ (unit digit is 4)

$5^3 = 125$ (unit digit is 5)

$6^3 = 216$ (unit digit is 6)

$7^3 = 343$ (unit digit is 3)

$8^3 = 512$ (unit digit is 2)

$9^3 = 729$ (unit digit is 9)

$10^3 = 1000$ (unit digit is 0)

Since the unit digit of 4913 is 3, the unit digit of its cube root must be 7 (because only the cube of a number ending in 7 ends in 3).

---

Step 3: Determine the tens digit of the cube root.

Consider the second group (the leftmost group), which is 4.

We need to find two consecutive perfect cubes between which this number lies.

$1^3 = 1$

$2^3 = 8$

We see that 4 lies between 1 and 8.

$1^3 < 4 < 2^3$

The tens digit of the cube root is the smaller number whose cube is in this range, which is the base of the smaller cube. Here, the smaller base is 1.

So, the tens digit of the cube root is 1.

---

Step 4: Combine the tens and unit digits.

Tens digit = 1

Unit digit = 7

Combining these digits, the estimated cube root of 4913 is 17.

---

We can verify this by calculating $17^3$:

$17^3 = 17 \times 17 \times 17 = 289 \times 17 = 4913$.

So, the estimate is correct.

To estimate the cube root of 17576, we follow these steps:

---

Step 1: Group the digits of the number from the right in sets of three.

Starting from the rightmost digit, we form groups of three digits. The number is 17576.

The first group from the right is $\mathbf{576}$.

The remaining digits form the second group, which is $\mathbf{17}$.

The groups are 17 and 576.

---

Step 2: Determine the unit digit of the cube root.

Consider the first group from the right, which is 576.

The unit digit of 576 is 6.

Now, we look at the unit digits of the cubes of natural numbers from 1 to 10:

$1^3 = 1$

$2^3 = 8$

$3^3 = 27$

$4^3 = 64$

$5^3 = 125$

$6^3 = 216$ (unit digit is 6)

$7^3 = 343$

$8^3 = 512$

$9^3 = 729$

$10^3 = 1000$

Since the unit digit of 17576 is 6, the unit digit of its cube root must be 6 (because only the cube of a number ending in 6 ends in 6).

---

Step 3: Determine the tens digit of the cube root.

Consider the second group (the leftmost group), which is 17.

We need to find two consecutive perfect cubes between which this number lies.

$1^3 = 1$

$2^3 = 8$

$3^3 = 27$

We see that 17 lies between 8 and 27.

$2^3 < 17 < 3^3$

The tens digit of the cube root is the smaller number whose cube is in this range, which is the base of the smaller cube. Here, the smaller base is 2.

So, the tens digit of the cube root is 2.

---

Step 4: Combine the tens and unit digits.

Tens digit = 2

Unit digit = 6

Combining these digits, the estimated cube root of 17576 is 26.

---

We can verify this by calculating $26^3$:

$26^3 = 26 \times 26 \times 26 = 676 \times 26 = 17576$.

So, the estimate is correct.

Let the length of the side of the cube be $s$ cm.

The formula for the volume ($V$) of a cube with side length $s$ is given by:

$V = s^3$

---

We are given that the volume of the cube is 729 cubic cm.

So, we have the equation:

---

To find the length of the side $s$, we need to find the cube root of 729.

$s = \sqrt[3]{729}$

---

We find the cube root of 729 by the prime factorization method.

Prime factorization of 729:

$\begin{array}{c|cc} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, the prime factorization of 729 is $3 \times 3 \times 3 \times 3 \times 3 \times 3$.

$729 = 3^6$

---

To find the cube root, we group the prime factors into triplets of equal factors.

$729 = (3 \times 3 \times 3) \times (3 \times 3 \times 3)$

$729 = 3^3 \times 3^3$

$729 = (3 \times 3)^3$

$729 = 9^3$

---

Now, we find the cube root:

$s = \sqrt[3]{729} = \sqrt[3]{9^3} = 9$

---

Therefore, the length of the side of the cube is 9 cm.

We need to find the cube of $0.3$, which is $(0.3)^3$.

---

The cube of a number is the number multiplied by itself three times.

$(0.3)^3 = 0.3 \times 0.3 \times 0.3$

---

First, multiply $0.3 \times 0.3$:

$0.3 \times 0.3 = 0.09$

Next, multiply the result by $0.3$ again:

$0.09 \times 0.3 = 0.027$

---

Alternatively, we can write the decimal as a fraction:

$0.3 = \frac{3}{10}$

Now, cube the fraction:

$\left(\frac{3}{10}\right)^3 = \frac{3^3}{10^3}$

$3^3 = 3 \times 3 \times 3 = 27$

$10^3 = 10 \times 10 \times 10 = 1000$

So, $\left(\frac{3}{10}\right)^3 = \frac{27}{1000}$.

---

Convert the fraction back to a decimal:

$\frac{27}{1000} = 0.027$

---

Therefore, the cube of $0.3$ is $0.027$.

To determine if 68600 is a perfect cube, we find its prime factorization.

---

Prime factorization of 68600:

$68600 = 686 \times 100$

Prime factorization of 686:

$\begin{array}{c|cc} 2 & 686 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

So, $686 = 2 \times 7 \times 7 \times 7 = 2^1 \times 7^3$.

Prime factorization of 100:

$\begin{array}{c|cc} 2 & 100 \\ \hline 2 & 50 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, $100 = 2 \times 2 \times 5 \times 5 = 2^2 \times 5^2$.

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Combining these factorizations for 68600:

$68600 = (2^1 \times 7^3) \times (2^2 \times 5^2)$

$68600 = 2^{1+2} \times 5^2 \times 7^3$

$68600 = 2^3 \times 5^2 \times 7^3$

---

For a number to be a perfect cube, the exponents of all its prime factors must be multiples of 3.

In the prime factorization of 68600, the exponents are:

Exponent of 2 is 3 (a multiple of 3).

Exponent of 5 is 2 (not a multiple of 3).

Exponent of 7 is 3 (a multiple of 3).

---

Since the exponent of the prime factor 5 is 2, which is not a multiple of 3, 68600 is not a perfect cube.

---

To make 68600 a perfect cube by multiplication, we need to make the exponent of 5 a multiple of 3. The smallest multiple of 3 greater than 2 is 3.

We have $5^2$. To make the exponent 3, we need to multiply by $5^{3-2} = 5^1 = 5$.

Multiplying 68600 by 5:

$68600 \times 5 = (2^3 \times 5^2 \times 7^3) \times 5^1 = 2^3 \times 5^{2+1} \times 7^3 = 2^3 \times 5^3 \times 7^3$

The new number is $(2 \times 5 \times 7)^3 = 70^3 = 343000$, which is a perfect cube.

---

The smallest number by which 68600 should be multiplied to make it a perfect cube is 5.

To find the unit digit of the cube root of 1331 without finding the actual cube root, we look at the unit digit of the number 1331.

---

The unit digit of 1331 is 1.

---

Now, we consider the unit digits of the cubes of the natural numbers from 0 to 9:

$0^3 = 0$ (Unit digit is 0)

$1^3 = 1$ (Unit digit is 1)

$2^3 = 8$ (Unit digit is 8)

$3^3 = 27$ (Unit digit is 7)

$4^3 = 64$ (Unit digit is 4)

$5^3 = 125$ (Unit digit is 5)

$6^3 = 216$ (Unit digit is 6)

$7^3 = 343$ (Unit digit is 3)

$8^3 = 512$ (Unit digit is 2)

$9^3 = 729$ (Unit digit is 9)

---

We observe which of these cubes has a unit digit of 1.

Only the cube of 1 ends in the digit 1 ($1^3 = 1$).

---

Therefore, the unit digit of the cube root of 1331 is 1.

(For verification, $11^3 = 11 \times 11 \times 11 = 121 \times 11 = 1331$. The cube root is 11, and its unit digit is 1).

We need to find the value of $\sqrt[3]{\frac{-512}{729}}$.

---

We use the properties of cube roots:

$\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}$ and $\sqrt[3]{-x} = -\sqrt[3]{x}$

Applying these properties, we have:

$\sqrt[3]{\frac{-512}{729}} = \frac{\sqrt[3]{-512}}{\sqrt[3]{729}} = \frac{-\sqrt[3]{512}}{\sqrt[3]{729}}$

---

Now, we find the cube root of the positive numbers 512 and 729 by prime factorization.

Prime factorization of 512:

$\begin{array}{c|cc} 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

So, $512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^9$.

Grouping factors in triplets: $512 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) = 2^3 \times 2^3 \times 2^3 = (2 \times 2 \times 2)^3 = 8^3$.

So, $\sqrt[3]{512} = 8$.

---

Prime factorization of 729:

$\begin{array}{c|cc} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6$.

Grouping factors in triplets: $729 = (3 \times 3 \times 3) \times (3 \times 3 \times 3) = 3^3 \times 3^3 = (3 \times 3)^3 = 9^3$.

So, $\sqrt[3]{729} = 9$.

---

Now, substitute the cube roots back into the expression for the cube root of the fraction:

$\sqrt[3]{\frac{-512}{729}} = \frac{-\sqrt[3]{512}}{\sqrt[3]{729}} = \frac{-8}{9}$

---

Therefore, the cube root of $\frac{-512}{729}$ is $\frac{-8}{9}$.

Here are the properties of the cubes of natural numbers relating to their unit digits and the number of zeros:

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Properties Related to Unit Digits:

The unit digit of the cube of a natural number is determined solely by the unit digit of the original number.

1. If the unit digit of a number is 0, 1, 4, 5, 6, or 9, the unit digit of its cube is the same digit.

Examples:

- Unit digit of 4 is 4. Cube of 4 is $4^3 = 64$. Unit digit of 64 is 4.

- Unit digit of 16 is 6. Cube of 16 is $16^3 = 4096$. Unit digit of 4096 is 6.

2. If the unit digit of a number is 2, the unit digit of its cube is 8.

Examples:

- Unit digit of 2 is 2. Cube of 2 is $2^3 = 8$. Unit digit of 8 is 8.

- Unit digit of 12 is 2. Cube of 12 is $12^3 = 1728$. Unit digit of 1728 is 8.

3. If the unit digit of a number is 8, the unit digit of its cube is 2.

Examples:

- Unit digit of 8 is 8. Cube of 8 is $8^3 = 512$. Unit digit of 512 is 2.

- Unit digit of 18 is 8. Cube of 18 is $18^3 = 5832$. Unit digit of 5832 is 2.

4. If the unit digit of a number is 3, the unit digit of its cube is 7.

Examples:

- Unit digit of 3 is 3. Cube of 3 is $3^3 = 27$. Unit digit of 27 is 7.

- Unit digit of 13 is 3. Cube of 13 is $13^3 = 2197$. Unit digit of 2197 is 7.

5. If the unit digit of a number is 7, the unit digit of its cube is 3.

Examples:

- Unit digit of 7 is 7. Cube of 7 is $7^3 = 343$. Unit digit of 343 is 3.

- Unit digit of 17 is 7. Cube of 17 is $17^3 = 4913$. Unit digit of 4913 is 3.

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Properties Related to the Number of Zeros:

The number of zeros at the end of a number affects the number of zeros at the end of its cube in a specific way.

1. If a natural number ends with $n$ zeros, then its cube ends with exactly $3n$ zeros.

Examples:

- The number 10 ends with 1 zero ($n=1$). Its cube is $10^3 = 1000$, which ends with 3 zeros ($3n = 3 \times 1 = 3$).

- The number 20 ends with 1 zero ($n=1$). Its cube is $20^3 = 8000$, which ends with 3 zeros ($3n = 3 \times 1 = 3$).

- The number 100 ends with 2 zeros ($n=2$). Its cube is $100^3 = 1000000$, which ends with 6 zeros ($3n = 3 \times 2 = 6$).

- The number 500 ends with 2 zeros ($n=2$). Its cube is $500^3 = 125000000$, which ends with 6 zeros ($3n = 3 \times 2 = 6$).

2. Consequently, a perfect cube must end with a number of zeros that is a multiple of 3 (i.e., 0, 3, 6, 9, etc., zeros). A number ending with 1, 2, 4, 5, 7, or 8 zeros cannot be a perfect cube.

Examples:

- The number 100 ends with 2 zeros. Since 2 is not a multiple of 3, 100 is not a perfect cube.

- The number 20000 ends with 4 zeros. Since 4 is not a multiple of 3, 20000 is not a perfect cube.

The prime factorization method for finding the cube root of a perfect cube involves the following steps:

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Step 1: Find the prime factorization of the given number. Express the number as a product of its prime factors.

Step 2: Group the prime factors into triplets (groups of three) of identical factors.

Step 3: From each group of three identical prime factors, take out one factor.

Step 4: Multiply the factors taken out from each group. This product will be the cube root of the original number.

---

Let's use this method to find the cube root of 5832.

Find $\sqrt[3]{5832}$

Step 1: Find the prime factorization of 5832.

$\begin{array}{c|cc} 2 & 5832 \\ \hline 2 & 2916 \\ \hline 2 & 1458 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, the prime factorization of 5832 is $2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$.

$5832 = 2^3 \times 3^6$

Step 2: Group the prime factors into triplets.

$5832 = (2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3)$

$5832 = 2^3 \times 3^3 \times 3^3$

Step 3: Take out one factor from each group.

From the triplet of 2s, take out 2.

From the first triplet of 3s, take out 3.

From the second triplet of 3s, take out 3.

Step 4: Multiply the factors taken out.

$\sqrt[3]{5832} = 2 \times 3 \times 3 = 18$

Therefore, the cube root of 5832 is 18.

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Now, let's use the method to find the cube root of 13824.

Find $\sqrt[3]{13824}$

Step 1: Find the prime factorization of 13824.

$\begin{array}{c|cc} 2 & 13824 \\ \hline 2 & 6912 \\ \hline 2 & 3456 \\ \hline 2 & 1728 \\ \hline 2 & 864 \\ \hline 2 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, the prime factorization of 13824 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$.

$13824 = 2^9 \times 3^3$

Step 2: Group the prime factors into triplets.

$13824 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)$

$13824 = 2^3 \times 2^3 \times 2^3 \times 3^3$

Step 3: Take out one factor from each group.

From the first triplet of 2s, take out 2.

From the second triplet of 2s, take out 2.

From the third triplet of 2s, take out 2.

From the triplet of 3s, take out 3.

Step 4: Multiply the factors taken out.

$\sqrt[3]{13824} = 2 \times 2 \times 2 \times 3 = 8 \times 3 = 24$

Therefore, the cube root of 13824 is 24.

To find the smallest number by which 13720 must be multiplied to get a perfect cube, we first find the prime factorization of 13720.

---

Prime factorization of 13720:

$13720 = 1372 \times 10$

Prime factorization of 10 is $2 \times 5$.

Prime factorization of 1372:

$\begin{array}{c|cc} 2 & 1372 \\ \hline 2 & 686 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

So, $1372 = 2 \times 2 \times 7 \times 7 \times 7 = 2^2 \times 7^3$.

---

Combining these factorizations for 13720:

$13720 = (2^2 \times 7^3) \times (2 \times 5)$

$13720 = 2^{2+1} \times 5^1 \times 7^3$

$13720 = 2^3 \times 5^1 \times 7^3$

---

For a number to be a perfect cube, the exponents of all its prime factors must be multiples of 3.

In the prime factorization of 13720, the exponents are:

Exponent of 2 is 3 (a multiple of 3).

Exponent of 5 is 1 (not a multiple of 3).

Exponent of 7 is 3 (a multiple of 3).

---

To make 13720 a perfect cube by multiplication, we need to make the exponent of 5 a multiple of 3. The smallest multiple of 3 greater than 1 is 3.

We have $5^1$. To make the exponent 3, we need to multiply by $5^{3-1} = 5^2 = 5 \times 5 = 25$.

The smallest number by which 13720 must be multiplied is 25.

---

Now, we find the number obtained after multiplying 13720 by 25:

New number = $13720 \times 25$

Using the prime factorization:

New number = $(2^3 \times 5^1 \times 7^3) \times 5^2 = 2^3 \times 5^{1+2} \times 7^3 = 2^3 \times 5^3 \times 7^3$

New number = $(2 \times 5 \times 7)^3 = 70^3$

New number = $70 \times 70 \times 70 = 4900 \times 70 = 343000$.

---

Finally, we find the cube root of the number obtained (343000).

From the prime factorization of the new number, which is $2^3 \times 5^3 \times 7^3$, we can find the cube root by taking one factor from each triplet:

$\sqrt[3]{343000} = \sqrt[3]{2^3 \times 5^3 \times 7^3} = 2 \times 5 \times 7 = 70$

The cube root of the number obtained is 70.

To find the smallest number by which 1188 must be divided to get a perfect cube, we first find the prime factorization of 1188.

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Prime factorization of 1188:

$\begin{array}{c|cc} 2 & 1188 \\ \hline 2 & 594 \\ \hline 3 & 297 \\ \hline 3 & 99 \\ \hline 3 & 33 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

So, the prime factorization of 1188 is $2 \times 2 \times 3 \times 3 \times 3 \times 11$.

$1188 = 2^2 \times 3^3 \times 11^1$

---

For a number to be a perfect cube, the exponents of all its prime factors in the prime factorization must be multiples of 3.

In the factorization $2^2 \times 3^3 \times 11^1$, the exponents are:

Exponent of 2 is 2 (not a multiple of 3).

Exponent of 3 is 3 (a multiple of 3).

Exponent of 11 is 1 (not a multiple of 3).

---

To make 1188 a perfect cube by dividing, we need to remove the prime factors whose exponents are not multiples of 3, or reduce their exponents to the largest multiple of 3 less than the current exponent (which is 0 in this case).

The prime factor 2 has an exponent of 2. We need to remove $2^2 = 2 \times 2 = 4$.

The prime factor 11 has an exponent of 1. We need to remove $11^1 = 11$.

The smallest number to divide by is the product of these excess prime factors: $2^2 \times 11^1 = 4 \times 11 = 44$.

The smallest number by which 1188 must be divided is 44.

---

Now, we find the number obtained after dividing 1188 by 44:

New number = $\frac{1188}{44}$

Using the prime factorization:

New number = $\frac{2^2 \times 3^3 \times 11^1}{2^2 \times 11^1} = 3^3$

New number = $27$.

The number obtained is 27, which is a perfect cube ($3^3$).

---

Finally, we find the cube root of the number obtained (27).

From the prime factorization of the new number, which is $3^3$, we can find the cube root:

$\sqrt[3]{27} = \sqrt[3]{3^3} = 3$

The cube root of the number obtained is 3.

The method of estimating the cube root of a number is a shortcut to quickly find the cube root of a perfect cube without performing prime factorization, especially useful for larger numbers. It relies on the patterns of unit digits and the magnitude of cubes.

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Here are the steps for estimating the cube root of a perfect cube:

Step 1: Group the digits. Start from the rightmost digit of the number and make groups of three digits. Place a comma or a bar after every three digits. The leftmost group might contain one, two, or three digits.

Step 2: Determine the unit digit of the cube root. Look at the unit digit of the first group from the right. Identify the digit from 0 to 9 whose cube has this same unit digit. This digit will be the unit digit of the cube root.

Step 3: Determine the tens digit of the cube root. Consider the leftmost group. Find two consecutive perfect cubes between which the value of this group lies. The base of the smaller of these two perfect cubes will be the tens digit of the cube root.

Step 4: Combine the digits. Combine the tens digit found in Step 3 and the unit digit found in Step 2 to get the estimated cube root of the number.

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Let's use this method to estimate the cube root of 91125.

Estimate $\sqrt[3]{91125}$

Step 1: Group the digits.

Group 91125 from the right in threes: $91 , 125$.

The groups are 125 (first group from right) and 91 (leftmost group).

Step 2: Determine the unit digit.

The unit digit of the first group (125) is 5.

We look for a digit whose cube ends in 5. We know $5^3 = 125$, which ends in 5.

So, the unit digit of $\sqrt[3]{91125}$ is 5.

Step 3: Determine the tens digit.

The leftmost group is 91.

We find consecutive perfect cubes: $4^3 = 64$ and $5^3 = 125$.

We see that 91 lies between 64 and 125 ($64 < 91 < 125$).

This is $4^3 < 91 < 5^3$. The smaller base is 4.

So, the tens digit of $\sqrt[3]{91125}$ is 4.

Step 4: Combine the digits.

Tens digit = 4, Unit digit = 5.

The estimated cube root of 91125 is 45.

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Now, let's use the method to estimate the cube root of 42875.

Estimate $\sqrt[3]{42875}$

Step 1: Group the digits.

Group 42875 from the right in threes: $42 , 875$.

The groups are 875 (first group from right) and 42 (leftmost group).

Step 2: Determine the unit digit.

The unit digit of the first group (875) is 5.

We look for a digit whose cube ends in 5. We know $5^3 = 125$, which ends in 5.

So, the unit digit of $\sqrt[3]{42875}$ is 5.

Step 3: Determine the tens digit.

The leftmost group is 42.

We find consecutive perfect cubes: $3^3 = 27$ and $4^3 = 64$.

We see that 42 lies between 27 and 64 ($27 < 42 < 64$).

This is $3^3 < 42 < 4^3$. The smaller base is 3.

So, the tens digit of $\sqrt[3]{42875}$ is 3.

Step 4: Combine the digits.

Tens digit = 3, Unit digit = 5.

The estimated cube root of 42875 is 35.

Method for finding the cube root of Decimals:

To find the cube root of a decimal, we can follow these steps:

1. Convert the decimal number into a fraction. The denominator will be a power of 10 (10, 100, 1000, etc.) corresponding to the number of decimal places.

2. Use the property of cube roots for fractions: $\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}$. Find the cube root of the numerator and the cube root of the denominator separately.

3. Simplify the resulting fraction.

4. Convert the resulting fraction back into a decimal form if required.

(Alternatively, count the number of decimal places in the original number. If it is a multiple of 3, say $3n$, find the cube root of the number ignoring the decimal point. The cube root will have $n$ decimal places.)

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Method for finding the cube root of Fractions:

To find the cube root of a fraction $\frac{a}{b}$, we use the property: $\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}$.

1. Find the cube root of the numerator ($\sqrt[3]{a}$) using methods like prime factorization or estimation.

2. Find the cube root of the denominator ($\sqrt[3]{b}$) using the same method.

3. Write the result as the fraction $\frac{\sqrt[3]{a}}{\sqrt[3]{b}}$.

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Let's find the cube root of the given numbers using these methods.

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(a) $\sqrt[3]{0.008}$

Convert 0.008 to a fraction: $0.008 = \frac{8}{1000}$.

Apply the fraction property:

$\sqrt[3]{0.008} = \sqrt[3]{\frac{8}{1000}} = \frac{\sqrt[3]{8}}{\sqrt[3]{1000}}$

Find the cube roots of the numerator and denominator:

$\sqrt[3]{8} = \sqrt[3]{2 \times 2 \times 2} = \sqrt[3]{2^3} = 2$

$\sqrt[3]{1000} = \sqrt[3]{10 \times 10 \times 10} = \sqrt[3]{10^3} = 10$

So, $\frac{\sqrt[3]{8}}{\sqrt[3]{1000}} = \frac{2}{10}$.

Simplify the fraction and convert back to decimal:

$\frac{2}{10} = \frac{1}{5} = 0.2$

Therefore, $\sqrt[3]{0.008} = \mathbf{0.2}$.

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(b) $\sqrt[3]{2.197}$

Convert 2.197 to a fraction: $2.197 = \frac{2197}{1000}$.

Apply the fraction property:

$\sqrt[3]{2.197} = \sqrt[3]{\frac{2197}{1000}} = \frac{\sqrt[3]{2197}}{\sqrt[3]{1000}}$

Find the cube roots of the numerator and denominator:

$\sqrt[3]{1000} = 10$ (from part a)

To find $\sqrt[3]{2197}$, we can estimate or use prime factorization. The unit digit is 7, so the cube root ends in 3. $10^3=1000$, $20^3=8000$. 2197 is between 1000 and 8000. Let's try 13. $13^3 = 13 \times 13 \times 13 = 169 \times 13 = 2197$. So, $\sqrt[3]{2197} = 13$.

So, $\frac{\sqrt[3]{2197}}{\sqrt[3]{1000}} = \frac{13}{10}$.

Convert the fraction back to decimal:

$\frac{13}{10} = 1.3$

Therefore, $\sqrt[3]{2.197} = \mathbf{1.3}$.

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(c) $\sqrt[3]{\frac{3375}{4096}}$

Apply the fraction property:

$\sqrt[3]{\frac{3375}{4096}} = \frac{\sqrt[3]{3375}}{\sqrt[3]{4096}}$

Find the cube root of the numerator (3375).

Using estimation: Unit digit is 5, so cube root ends in 5. Leftmost group is 3 ($1^3=1, 2^3=8$). $1^3 < 3 < 2^3$, so tens digit is 1. Estimate is 15. $15^3 = 3375$. So, $\sqrt[3]{3375} = 15$.

Find the cube root of the denominator (4096).

Using estimation: Unit digit is 6, so cube root ends in 6. Leftmost group is 4 ($1^3=1, 2^3=8$). $1^3 < 4 < 2^3$, so tens digit is 1. Estimate is 16. $16^3 = 4096$. So, $\sqrt[3]{4096} = 16$.

Substitute the cube roots:

$\frac{\sqrt[3]{3375}}{\sqrt[3]{4096}} = \frac{15}{16}$

Therefore, $\sqrt[3]{\frac{3375}{4096}} = \mathbf{\frac{15}{16}}$.

Let the length of each side of the cubical box be $s$ cm.

The volume ($V$) of a cube with side length $s$ is given by the formula:

$V = s^3$

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We are given that the volume of the cubical box is 13824 cubic cm.

So, we have the equation:

To find the length of the side $s$, we need to find the cube root of 13824.

$s = \sqrt[3]{13824}$

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We find the cube root of 13824 using the prime factorization method.

Prime factorization of 13824:

$\begin{array}{c|cc} 2 & 13824 \\ \hline 2 & 6912 \\ \hline 2 & 3456 \\ \hline 2 & 1728 \\ \hline 2 & 864 \\ \hline 2 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The prime factorization of 13824 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$.

We group the prime factors into triplets:

$13824 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)$

$13824 = 2^3 \times 2^3 \times 2^3 \times 3^3$

Take one factor from each triplet:

$\sqrt[3]{13824} = 2 \times 2 \times 2 \times 3 = 8 \times 3 = 24$

So, $s = 24$ cm.

The length of each side of the box is 24 cm.

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Next, we need to find the amount of wire needed to go along all the edges of the box.

A cube has 12 edges, and all edges are of equal length.

The length of each edge is $s = 24$ cm.

The total length of all edges is the sum of the lengths of the 12 edges.

Total length of wire = $12 \times s$

Total length of wire = $12 \times 24$

Let's calculate the product:

$\begin{array}{cc}& & 2 & 4 \\ \times & & 1 & 2 \\ \hline && 4 & 8 \\ & 2 & 4 & \times \\ \hline & 2 & 8 & 8 \\ \hline \end{array}$

So, $12 \times 24 = 288$ cm.

The amount of wire needed is 288 cm.

Here are the properties used for finding the cube root of a product and a quotient:

Cube Root of a Product Property: The cube root of a product of two numbers is equal to the product of their cube roots. Mathematically, for any numbers $a$ and $b$, $\sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b}$.

Cube Root of a Quotient Property: The cube root of a quotient of two numbers is equal to the quotient of their cube roots. Mathematically, for any number $a$ and a non-zero number $b$, $\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}$.

Also, for a negative number $-x$, its cube root is the negative of the cube root of the positive number $x$. Mathematically, $\sqrt[3]{-x} = -\sqrt[3]{x}$.

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Let's evaluate $\sqrt[3]{-125 \times 729}$.

We can use the cube root of a product property: $\sqrt[3]{-125 \times 729} = \sqrt[3]{-125} \times \sqrt[3]{729}$.

Now, we find the cube root of each number:

Find $\sqrt[3]{-125}$. Using the property for negative numbers, $\sqrt[3]{-125} = -\sqrt[3]{125}$.

Since $5 \times 5 \times 5 = 125$, we have $\sqrt[3]{125} = 5$.

So, $\sqrt[3]{-125} = -5$.

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Find $\sqrt[3]{729}$. We can find the prime factorization of 729:

$\begin{array}{c|cc} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6$.

Grouping in triplets: $729 = (3 \times 3 \times 3) \times (3 \times 3 \times 3) = 3^3 \times 3^3 = (3 \times 3)^3 = 9^3$.

So, $\sqrt[3]{729} = 9$.

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Now, multiply the cube roots:

$\sqrt[3]{-125 \times 729} = \sqrt[3]{-125} \times \sqrt[3]{729} = (-5) \times 9 = -45$.

Therefore, $\sqrt[3]{-125 \times 729} = \mathbf{-45}$.

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Let's evaluate $\sqrt[3]{\frac{-512}{343}}$.

We use the cube root of a quotient property: $\sqrt[3]{\frac{-512}{343}} = \frac{\sqrt[3]{-512}}{\sqrt[3]{343}}$.

Now, we find the cube root of the numerator and the denominator.

Find $\sqrt[3]{-512}$. Using the property for negative numbers, $\sqrt[3]{-512} = -\sqrt[3]{512}$.

We find the prime factorization of 512:

$\begin{array}{c|cc} 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

So, $512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^9$.

Grouping in triplets: $512 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) = 2^3 \times 2^3 \times 2^3 = (2 \times 2 \times 2)^3 = 8^3$.

So, $\sqrt[3]{512} = 8$.

Therefore, $\sqrt[3]{-512} = -8$.

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Find $\sqrt[3]{343}$. We find the prime factorization of 343:

$\begin{array}{c|cc} 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

So, $343 = 7 \times 7 \times 7 = 7^3$.

So, $\sqrt[3]{343} = 7$.

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Now, form the quotient of the cube roots:

$\sqrt[3]{\frac{-512}{343}} = \frac{\sqrt[3]{-512}}{\sqrt[3]{343}} = \frac{-8}{7}$.

Therefore, $\sqrt[3]{\frac{-512}{343}} = \mathbf{\frac{-8}{7}}$.

We need to find the cube root of $1728 \times 27$ and verify the property $\sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b}$ with $a=1728$ and $b=27$.

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Step 1: Find the cube root of $1728$.

First, find the prime factorization of $1728$.

$\begin{array}{c|cc} 2 & 1728 \\ \hline 2 & 864 \\ \hline 2 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^6 \times 3^3 = (2^2)^3 \times 3^3 = (2^2 \times 3)^3 = (4 \times 3)^3 = 12^3$.

Therefore, $\sqrt[3]{1728} = \sqrt[3]{12^3} = 12$.

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Step 2: Find the cube root of $27$.

Find the prime factorization of $27$.

$\begin{array}{c|cc} 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $27 = 3 \times 3 \times 3 = 3^3$.

Therefore, $\sqrt[3]{27} = \sqrt[3]{3^3} = 3$.

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Step 3: Calculate $\sqrt[3]{a} \times \sqrt[3]{b}$ for $a=1728$ and $b=27$.

$\sqrt[3]{1728} \times \sqrt[3]{27} = 12 \times 3 = 36$.

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Step 4: Calculate the product $a \times b = 1728 \times 27$.

$1728 \times 27$

$\begin{array}{cc}& & 1 & 7 & 2 & 8 \\ \times & & & & 2 & 7 \\ \hline && 1 & 2 & 0 & 9 & 6 \\ & 3 & 4 & 5 & 6 & \times \\ \hline 4 & 6 & 6 & 5 & 6 \\ \hline \end{array}$

$1728 \times 27 = 46656$.

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Step 5: Find the cube root of the product $1728 \times 27 = 46656$.

Find the prime factorization of $46656$.

We know $1728 = 2^6 \times 3^3$ and $27 = 3^3$.

So, $1728 \times 27 = (2^6 \times 3^3) \times 3^3 = 2^6 \times 3^{3+3} = 2^6 \times 3^6 = (2^2)^3 \times (3^2)^3 = (2^2 \times 3^2)^3 = (4 \times 9)^3 = 36^3$.

Therefore, $\sqrt[3]{46656} = \sqrt[3]{36^3} = 36$.

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Step 6: Verify the property $\sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b}$.

From Step 5, $\sqrt[3]{1728 \times 27} = \sqrt[3]{46656} = 36$.

From Step 3, $\sqrt[3]{1728} \times \sqrt[3]{27} = 36$.

Since both results are equal to $36$, the property $\sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b}$ is verified for $a=1728$ and $b=27$.

Given: The volume of the cube is $0.001728$ cubic meters.

To Find: The length of the side of the cube in centimeters.

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Solution:

Let the length of the side of the cube be $s$ meters.

The volume $V$ of a cube is given by the formula $V = s^3$.

We are given that the volume is $0.001728 \text{ m}^3$.

To find the side length $s$, we take the cube root of the volume:

We can write the decimal $0.001728$ as a fraction:

So, the side length in meters is:

Using the property $\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}$, we have:

We find the cube root of the numerator and the denominator:

$\sqrt[3]{1728} = 12$ (as calculated in the previous question)

$\sqrt[3]{1000000} = \sqrt[3]{100^3} = 100$

Substituting these values:

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The side length is $0.12$ meters. We need to convert this length to centimeters.

We know that $1$ meter is equal to $100$ centimeters.

To convert meters to centimeters, we multiply the length in meters by $100$.

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Thus, the length of the side of the cube is $12$ centimeters.

We need to find the cube root of the mixed fraction $3\frac{3}{8}$.

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Step 1: Convert the mixed fraction to an improper fraction.

The mixed fraction is $3\frac{3}{8}$.

To convert a mixed fraction $a\frac{b}{c}$ to an improper fraction, we use the formula $\frac{(a \times c) + b}{c}$.

The improper fraction is $\frac{27}{8}$.

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Step 2: Find the cube root of the improper fraction.

We need to find $\sqrt[3]{\frac{27}{8}}$.

The cube root of a fraction is the cube root of the numerator divided by the cube root of the denominator:

So, $\sqrt[3]{\frac{27}{8}} = \frac{\sqrt[3]{27}}{\sqrt[3]{8}}$.

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Step 3: Find the cube root of the numerator ($27$).

We need a number that, when multiplied by itself three times, equals $27$.

We know that $3 \times 3 \times 3 = 27$.

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Step 4: Find the cube root of the denominator ($8$).

We need a number that, when multiplied by itself three times, equals $8$.

We know that $2 \times 2 \times 2 = 8$.

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Step 5: Combine the cube roots.

Now, substitute the cube roots back into the expression:

The cube root of $3\frac{3}{8}$ is $\frac{3}{2}$.

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The cube root of $3\frac{3}{8}$ is $\frac{3}{2}$, which can also be written as the mixed number $1\frac{1}{2}$ or the decimal $1.5$.

The answer is $\frac{3}{2}$.

The cube of a number is the result of multiplying the number by itself three times. If a number is $x$, its cube is $x \times x \times x$, denoted as $x^3$.

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Cube of a Negative Integer:

A negative integer is an integer less than zero (e.g., $-1, -2, -3, ...$).

When we cube a negative integer, we multiply it by itself three times:

Let's consider the sign of the result:

The product of the first two negative numbers is positive:

Then, we multiply this positive result by the third negative number:

Therefore, the cube of a negative integer is always a negative integer.

Example 1: Find the cube of $-5$.

So, the cube of $-5$ is $-125$.

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Cube of a Negative Rational Number:

A negative rational number is a number that can be expressed in the form $-\frac{p}{q}$, where $p$ and $q$ are positive integers and $q \neq 0$.

When we cube a negative rational number, we multiply it by itself three times:

Similar to the negative integer case, we consider the signs and the multiplication of fractions:

Therefore, the cube of a negative rational number is always a negative rational number.

Example 2: Find the cube of $-\frac{2}{3}$.

So, the cube of $-\frac{2}{3}$ is $-\frac{8}{27}$.

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Relationship between the Sign of the Cube and the Sign of the Base:

From the examples and the explanations above, we can see a clear relationship between the sign of the base number and the sign of its cube:

• If the base number is positive, its cube is positive ($+ \times + \times + = +$).
• If the base number is negative, its cube is negative ($- \times - \times - = -$).

In summary, the sign of the cube of a number is always the same as the sign of the number itself.