Chapter 7 Cubes and Cube Roots

A number is a perfect cube if its prime factors can be grouped into triplets of equal factors.

To determine if 243 is a perfect cube, we find its prime factorisation.

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Prime factorisation of 243:

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

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So, the prime factorisation of 243 is $3 \times 3 \times 3 \times 3 \times 3$.

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Now, we group the prime factors into triplets:

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In the prime factorisation of 243, we have one triplet of 3s ($3^3$) and two factors of 3 left ($3^2$) which do not form a complete triplet.

For a number to be a perfect cube, all prime factors must appear in groups of three.

Since the prime factor 3 does not form a complete triplet, 243 is not a perfect cube.

No, 243 is not a perfect cube.

To check if 392 is a perfect cube, we find its prime factorisation and group the factors in triplets.

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Let's find the prime factorisation of 392.

$\begin{array}{c|cc} 2 & 392 \\ \hline 2 & 196 \\ \hline 2 & 98 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

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The prime factorisation of 392 is $2 \times 2 \times 2 \times 7 \times 7$.

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Now, let's group the prime factors into triplets:

In the prime factorisation of 392, the factor 2 appears as a triplet ($2^3$), but the factor 7 appears only twice ($7^2$). For 392 to be a perfect cube, all prime factors must appear in groups of three.

Since the prime factor 7 does not form a complete triplet, 392 is not a perfect cube.

No, 392 is not a perfect cube.

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To make 392 a perfect cube, we need to multiply it by the smallest natural number that will complete the triplet of the prime factor 7.

The prime factorisation is $2^3 \times 7^2$. We need one more factor of 7 to make $7^2$ into $7^3$.

The smallest natural number required is 7.

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Multiplying 392 by 7:

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

The product 2744 is a perfect cube ($14^3$).

The smallest natural number by which 392 must be multiplied to get a perfect cube is 7.

To check if 53240 is a perfect cube, we find its prime factorisation and group the factors in triplets.

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Let's find the prime factorisation of 53240.

$\begin{array}{c|cc} 2 & 53240 \\ \hline 2 & 26620 \\ \hline 2 & 13310 \\ \hline 5 & 6655 \\ \hline 11 & 1331 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

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The prime factorisation of 53240 is $2 \times 2 \times 2 \times 5 \times 11 \times 11 \times 11$.

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Now, let's group the prime factors into triplets:

In the prime factorisation of 53240, the factors 2 and 11 appear as triplets ($2^3$ and $11^3$), but the factor 5 appears only once ($5^1$). For 53240 to be a perfect cube, all prime factors must appear in groups of three.

Since the prime factor 5 does not form a complete triplet, 53240 is not a perfect cube.

No, 53240 is not a perfect cube.

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To make 53240 a perfect cube by division, we need to divide it by the smallest natural number that will remove the prime factors which are not part of a triplet.

The prime factorisation is $2^3 \times 5^1 \times 11^3$. The prime factor that is not part of a triplet is $5^1$.

To make the quotient a perfect cube, we must divide 53240 by 5.

The smallest natural number required is 5.

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Dividing 53240 by 5:

$\begin{array}{r} 10648 \\ 5{\overline{\smash{\big)}\,53240\phantom{)}}} \\ \underline{-~\phantom{()}(5)\phantom{240)}} \\ 03\phantom{3240} \\ \underline{-~\phantom{()}(0)\phantom{240)}} \\ 32\phantom{040} \\ \underline{-~\phantom{()}(30)\phantom{040)}} \\ 24\phantom{40} \\ \underline{-~\phantom{()}(20)\phantom{00)}} \\ 40 \\ \underline{-~\phantom{()}(40)} \\ 0 \end{array}$

Let's check the prime factorisation of the quotient 10648:

The quotient 10648 is a perfect cube ($22^3$).

The smallest natural number by which 53240 must be divided so that the quotient is a perfect cube is 5.

To check if 1188 is a perfect cube, we find its prime factorisation and group the factors in triplets.

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Let's find the prime factorisation of 1188.

$\begin{array}{c|cc} 2 & 1188 \\ \hline 2 & 594 \\ \hline 3 & 297 \\ \hline 3 & 99 \\ \hline 3 & 33 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

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The prime factorisation of 1188 is $2 \times 2 \times 3 \times 3 \times 3 \times 11$.

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Now, let's group the prime factors into triplets:

In the prime factorisation of 1188, the factor 3 appears as a triplet ($3^3$), but the factors 2 and 11 do not appear in complete triplets ($2^2$ and $11^1$). For 1188 to be a perfect cube, all prime factors must appear in groups of three.

Since the prime factors 2 and 11 do not form complete triplets, 1188 is not a perfect cube.

No, 1188 is not a perfect cube.

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To make 1188 a perfect cube by division, we need to divide it by the smallest natural number that will remove the prime factors which are not part of a triplet.

The prime factorisation is $2^2 \times 3^3 \times 11^1$. The prime factors that are not part of a triplet are $2^2$ and $11^1$.

To make the quotient a perfect cube, we must divide 1188 by the product of these factors: $2^2 \times 11 = 4 \times 11 = 44$.

The smallest natural number required is 44.

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Dividing 1188 by 44:

Let's check the prime factorisation of the quotient 27:

The quotient 27 is a perfect cube ($3^3$).

The smallest natural number by which 1188 must be divided so that the quotient is a perfect cube is 44.

To check if 68600 is a perfect cube, we find its prime factorisation and group the factors in triplets.

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Let's find the prime factorisation of 68600.

$\begin{array}{c|cc} 2 & 68600 \\ \hline 2 & 34300 \\ \hline 2 & 17150 \\ \hline 5 & 8575 \\ \hline 5 & 1715 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

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The prime factorisation of 68600 is $2 \times 2 \times 2 \times 5 \times 5 \times 7 \times 7 \times 7$.

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Now, let's group the prime factors into triplets:

In the prime factorisation of 68600, the factors 2 and 7 appear as triplets ($2^3$ and $7^3$), but the factor 5 appears only twice ($5^2$). For 68600 to be a perfect cube, all prime factors must appear in groups of three.

Since the prime factor 5 does not form a complete triplet, 68600 is not a perfect cube.

No, 68600 is not a perfect cube.

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To make 68600 a perfect cube by multiplication, we need to multiply it by the smallest natural number that will complete the triplet of the prime factor 5.

The prime factorisation is $2^3 \times 5^2 \times 7^3$. We need one more factor of 5 to make $5^2$ into $5^3$.

The smallest natural number required is 5.

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Multiplying 68600 by 5:

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

The product 343000 is a perfect cube ($70^3$).

The smallest natural number by which 68600 must be multiplied to get a perfect cube is 5.

To check if a number is a perfect cube, we find its prime factorisation and group the factors into triplets. If all prime factors form complete triplets, then the number is a perfect cube; otherwise, it is not.

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(i) 216:

Prime factorisation of 216:

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

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

Since all prime factors (2 and 3) form complete triplets, 216 is a perfect cube.

216 is a perfect cube.

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(ii) 128:

Prime factorisation of 128:

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

$128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^7$

$128 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times 2 = 2^3 \times 2^3 \times 2^1$

In the prime factorisation of 128, the factor 2 does not appear in complete triplets (there is one factor of 2 remaining). Therefore, 128 is not a perfect cube.

128 is not a perfect cube.

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(iii) 1000:

Prime factorisation of 1000:

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

$1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 = 2^3 \times 5^3 = (2 \times 5)^3 = 10^3$

Since all prime factors (2 and 5) form complete triplets, 1000 is a perfect cube.

1000 is a perfect cube.

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(iv) 100:

Prime factorisation of 100:

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

$100 = 2 \times 2 \times 5 \times 5 = 2^2 \times 5^2$

In the prime factorisation of 100, the factors 2 and 5 do not appear in complete triplets. Therefore, 100 is not a perfect cube.

100 is not a perfect cube.

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(v) 46656:

Prime factorisation of 46656:

$\begin{array}{c|cc} 2 & 46656 \\ \hline 2 & 23328 \\ \hline 2 & 11664 \\ \hline 2 & 5832 \\ \hline 2 & 2916 \\ \hline 2 & 1458 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$46656 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 2^6 \times 3^6$

$46656 = (2^3) \times (2^3) \times (3^3) \times (3^3) = (2 \times 2 \times 3 \times 3)^3 = (36)^3$

Since all prime factors (2 and 3) form complete triplets, 46656 is a perfect cube.

46656 is a perfect cube.

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The numbers that are not perfect cubes are 128 and 100.

To find the smallest number by which a given number must be multiplied to obtain a perfect cube, we find the prime factorisation of the number and identify the factors that are not in triplets. The required smallest number is the product of the factors needed to complete the triplets.

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(i) 243:

Prime factorisation of 243:

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

$243 = 3 \times 3 \times 3 \times 3 \times 3 = 3^5 = 3^3 \times 3^2$

The factor 3 is not in a complete triplet (we have $3^2$). To make it a triplet, we need one more factor of 3 ($3^1$).

Smallest number to multiply = 3.

The smallest number by which 243 must be multiplied to obtain a perfect cube is 3.

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(ii) 256:

Prime factorisation of 256:

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

$256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^8 = 2^3 \times 2^3 \times 2^2$

The factor 2 is not in a complete triplet (we have $2^2$). To make it a triplet, we need one more factor of 2 ($2^1$).

Smallest number to multiply = 2.

The smallest number by which 256 must be multiplied to obtain a perfect cube is 2.

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(iii) 72:

Prime factorisation of 72:

$\begin{array}{c|cc} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2$

The factor 3 is not in a complete triplet (we have $3^2$). To make it a triplet, we need one more factor of 3 ($3^1$).

Smallest number to multiply = 3.

The smallest number by which 72 must be multiplied to obtain a perfect cube is 3.

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(iv) 675:

Prime factorisation of 675:

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

$675 = 3 \times 3 \times 3 \times 5 \times 5 = 3^3 \times 5^2$

The factor 5 is not in a complete triplet (we have $5^2$). To make it a triplet, we need one more factor of 5 ($5^1$).

Smallest number to multiply = 5.

The smallest number by which 675 must be multiplied to obtain a perfect cube is 5.

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(v) 100:

Prime factorisation of 100:

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

$100 = 2 \times 2 \times 5 \times 5 = 2^2 \times 5^2$

The factors 2 and 5 are not in complete triplets (we have $2^2$ and $5^2$). To make them triplets, we need one more factor of 2 ($2^1$) and one more factor of 5 ($5^1$).

Smallest number to multiply = $2 \times 5 = 10$.

The smallest number by which 100 must be multiplied to obtain a perfect cube is 10.

To find the smallest number by which a given number must be divided to obtain a perfect cube, we find the prime factorisation of the number and identify the factors that are not part of a complete triplet. The required smallest number is the product of these excess factors.

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(i) 81:

Prime factorisation of 81:

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

$81 = 3 \times 3 \times 3 \times 3 = 3^4 = 3^3 \times 3^1$

The factor 3 is not part of a complete triplet (we have one factor of 3 left over after forming a triplet). To make the quotient a perfect cube, we must divide by this excess factor ($3^1$).

Smallest number to divide = 3.

The smallest number by which 81 must be divided to obtain a perfect cube is 3. (Quotient = $81 \div 3 = 27 = 3^3$)

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(ii) 128:

Prime factorisation of 128:

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

$128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^7 = 2^3 \times 2^3 \times 2^1$

The factor 2 is not part of a complete triplet (we have one factor of 2 left over after forming triplets). To make the quotient a perfect cube, we must divide by this excess factor ($2^1$).

Smallest number to divide = 2.

The smallest number by which 128 must be divided to obtain a perfect cube is 2. (Quotient = $128 \div 2 = 64 = 4^3$)

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(iii) 135:

Prime factorisation of 135:

$\begin{array}{c|cc} 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$135 = 3 \times 3 \times 3 \times 5 = 3^3 \times 5^1$

The factor 5 is not part of a complete triplet (we have one factor of 5). To make the quotient a perfect cube, we must divide by this excess factor ($5^1$).

Smallest number to divide = 5.

The smallest number by which 135 must be divided to obtain a perfect cube is 5. (Quotient = $135 \div 5 = 27 = 3^3$)

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(iv) 192:

Prime factorisation of 192:

$\begin{array}{c|cc} 2 & 192 \\ \hline 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^6 \times 3^1 = (2^3) \times (2^3) \times 3^1$

The factor 3 is not part of a complete triplet (we have one factor of 3). To make the quotient a perfect cube, we must divide by this excess factor ($3^1$).

Smallest number to divide = 3.

The smallest number by which 192 must be divided to obtain a perfect cube is 3. (Quotient = $192 \div 3 = 64 = 4^3$)

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(v) 704:

Prime factorisation of 704:

$\begin{array}{c|cc} 2 & 704 \\ \hline 2 & 352 \\ \hline 2 & 176 \\ \hline 2 & 88 \\ \hline 2 & 44 \\ \hline 2 & 22 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

$704 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11 = 2^6 \times 11^1 = (2^3) \times (2^3) \times 11^1$

The factor 11 is not part of a complete triplet (we have one factor of 11). To make the quotient a perfect cube, we must divide by this excess factor ($11^1$).

Smallest number to divide = 11.

The smallest number by which 704 must be divided to obtain a perfect cube is 11. (Quotient = $704 \div 11 = 64 = 4^3$)

Given:

The dimensions of the plasticine cuboid are 5 cm, 2 cm, and 5 cm.

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To Find:

The number of such cuboids required to form a perfect cube.

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Solution:

The volume of the given cuboid is calculated by multiplying its dimensions:

To form a perfect cube using multiple identical cuboids, the total volume of the combined solid must be a perfect cube number.

A perfect cube number is a number whose prime factors, when written in prime factorisation form, all have powers that are multiples of 3.

Let the number of cuboids needed to form a perfect cube be $n$.

The total volume of $n$ cuboids will be $n \times (\text{Volume of one cuboid})$.

For the Total Volume to be a perfect cube, the powers of the prime factors 2 and 5 in the expression $n \times 2^1 \times 5^2$ must be multiples of 3.

Let the prime factorisation of $n$ be $2^a \times 5^b \times \dots$, where the dots represent other prime factors (which must have powers that are multiples of 3 for the smallest $n$).

For Total Volume to be a perfect cube, the exponents in (iii) must be multiples of 3.

For the smallest $n$, we need the smallest non-negative integers $a$ and $b$ such that $a+1$ and $b+2$ are the smallest possible multiples of 3 (at least 1 and 2, respectively). The smallest multiple of 3 greater than or equal to 1 is 3. The smallest multiple of 3 greater than or equal to 2 is 3.

So, we set $a+1 = 3 \implies a = 2$.

And $b+2 = 3 \implies b = 1$.

No other prime factors are needed in the smallest $n$, so the dots represent factors with power 0.

Thus, Parikshit needs 20 such cuboids.

If he uses 20 cuboids, the total volume will be:

The total volume is $10^3 = 1000$ cubic cm, which is a perfect cube with side length 10 cm.

Therefore, Parikshit will need 20 cuboids to form a cube.

To find the cube root of a number using the prime factorisation method, we find the prime factorisation of the number and group the factors into triplets. The cube root is the product of one factor from each triplet.

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Let's find the prime factorisation of 8000.

$\begin{array}{c|cc} 2 & 8000 \\ \hline 2 & 4000 \\ \hline 2 & 2000 \\ \hline 2 & 1000 \\ \hline 2 & 500 \\ \hline 2 & 250 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

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The prime factorisation of 8000 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5$.

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Now, we group the prime factors into triplets:

To find the cube root, we take one factor from each triplet:

The cube root of 8000 is 20.

To find the cube root of 13824 by the prime factorisation method, we first find the prime factorisation of 13824 and then group the factors into triplets.

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Let's find the prime factorisation of 13824.

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

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The prime factorisation of 13824 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$.

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Now, we group the prime factors into triplets:

To find the cube root, we take one factor from each triplet:

The cube root of 13824 is 24.

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

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Step 1: Group the digits.

Starting from the rightmost digit of 17576, we group the digits in sets of three. We get two groups: 17 and 576.

The first group is 576, and the second group is 17.

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

Look at the first group from the right, which is 576. The unit digit of this group is 6.

Now, consider the cubes of numbers from 1 to 9 and their unit digits:

• $1^3 = 1$ (unit digit 1)
• $2^3 = 8$ (unit digit 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 unit digit of the group 576 is 6. The unit digit of the cube root will be the unit digit of the number whose cube has a unit digit of 6. From the list above, only $6^3 = 216$ has a unit digit of 6. So, the unit digit of the cube root of 17576 is 6.

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

Now, consider the second group from the right, which is 17.

We need to find two consecutive perfect cubes between which 17 lies.

• $2^3 = 8$
• $3^3 = 27$

We see that $8 < 17 < 27$, which means $2^3 < 17 < 3^3$.

The smaller of these two numbers is 2.

The tens digit of the cube root will be this smaller number, which is 2.

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Combining the tens digit and the unit digit, the estimated cube root is 26.

Let's check by cubing 26:

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

$26 \times 26 = 676$

$676 \times 26$

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

$26^3 = 17576$

The estimated cube root is correct.

The cube root of 17576 is 26.

To find the cube root of a number by prime factorisation, we find the prime factorisation of the number, group the identical factors into triplets, and take one factor from each triplet. The product of these factors is the cube root.

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(i) 64:

Prime factorisation of 64:

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

$64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) = 2^3 \times 2^3 \ $$ = (2 \times 2)^3 = 4^3$

$\sqrt[3]{64} = 2 \times 2 = 4$

The cube root of 64 is 4.

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(ii) 512:

Prime factorisation of 512:

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

$512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \ $$= (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \ $$ = 2^3 \times 2^3 \times 2^3 = (2 \times 2 \times 2)^3 = 8^3$

$\sqrt[3]{512} = 2 \times 2 \times 2 = 8$

The cube root of 512 is 8.

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(iii) 10648:

Prime factorisation of 10648:

$\begin{array}{c|cc} 2 & 10648 \\ \hline 2 & 5324 \\ \hline 2 & 2662 \\ \hline 11 & 1331 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

$10648 = 2 \times 2 \times 2 \times 11 \times 11 \times 11 = (2 \times 2 \times 2) \times (11 \times 11 \times 11) \ $$ = 2^3 \times 11^3 = (2 \times 11)^3 = 22^3$

$\sqrt[3]{10648} = 2 \times 11 = 22$

The cube root of 10648 is 22.

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(iv) 27000:

Prime factorisation of 27000:

$\begin{array}{c|cc} 2 & 27000 \\ \hline 2 & 13500 \\ \hline 2 & 6750 \\ \hline 3 & 3375 \\ \hline 3 & 1125 \\ \hline 3 & 375 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$27000 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 \ $$ = (2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (5 \times 5 \times 5) \ $$ = 2^3 \times 3^3 \times 5^3 = (2 \times 3 \times 5)^3 = 30^3$

$\sqrt[3]{27000} = 2 \times 3 \times 5 = 30$

The cube root of 27000 is 30.

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(v) 15625:

Prime factorisation of 15625:

$\begin{array}{c|cc} 5 & 15625 \\ \hline 5 & 3125 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$15625 = 5 \times 5 \times 5 \times 5 \times 5 \times 5 = (5 \times 5 \times 5) \times (5 \times 5 \times 5) = 5^3 \times 5^3 \ $$ = (5 \times 5)^3 = 25^3$

$\sqrt[3]{15625} = 5 \times 5 = 25$

The cube root of 15625 is 25.

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(vi) 13824:

Prime factorisation of 13824:

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

$13824 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \ $$ = (2^3) \times (2^3) \times (2^3) \times (3^3) = (2 \times 2 \times 2 \times 3)^3 = 24^3$

$\sqrt[3]{13824} = 2 \times 2 \times 2 \times 3 = 24$

The cube root of 13824 is 24.

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(vii) 110592:

Prime factorisation of 110592:

$\begin{array}{c|cc} 2 & 110592 \\ \hline 2 & 55296 \\ \hline 2 & 27648 \\ \hline 2 & 13824 \\ \hline 2 & 6912 \\ \hline 2 & 3456 \\ \hline 2 & 1728 \\ \hline 2 & 864 \\ \hline 2 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$110592 = 2^{12} \times 3^3 = (2^4)^3 \times 3^3 = (16 \times 3)^3 = 48^3$

$\sqrt[3]{110592} = 2^4 \times 3 = 16 \times 3 = 48$.

The cube root of 110592 is 48.

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(viii) 46656:

Prime factorisation of 46656:

$\begin{array}{c|cc} 2 & 46656 \\ \hline 2 & 23328 \\ \hline 2 & 11664 \\ \hline 2 & 5832 \\ \hline 2 & 2916 \\ \hline 2 & 1458 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$46656 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \ $$ = (2^3) \times (2^3) \times (3^3) \times (3^3) = (2 \times 2 \times 3 \times 3)^3 = 36^3$

$\sqrt[3]{46656} = 2 \times 2 \times 3 \times 3 = 36$

The cube root of 46656 is 36.

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(ix) 175616:

Prime factorisation of 175616:

$\begin{array}{c|cc} 2 & 175616 \\ \hline 2 & 87808 \\ \hline 2 & 43904 \\ \hline 2 & 21952 \\ \hline 2 & 10976 \\ \hline 2 & 5488 \\ \hline 2 & 2744 \\ \hline 2 & 1372 \\ \hline 2 & 686 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

$175616 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7 \times 7 \times 7 \ $$ = (2^3) \times (2^3) \times (2^3) \times (7^3) = (2 \times 2 \times 2 \times 7)^3 = 56^3$

$\sqrt[3]{175616} = 2 \times 2 \times 2 \times 7 = 56$

The cube root of 175616 is 56.

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(x) 91125:

Prime factorisation of 91125:

$\begin{array}{c|cc} 3 & 91125 \\ \hline 3 & 30375 \\ \hline 3 & 10125 \\ \hline 3 & 3375 \\ \hline 3 & 1125 \\ \hline 3 & 375 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$91125 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 \ $$ = (3^3) \times (3^3) \times (5^3) = (3 \times 3 \times 5)^3 = 45^3$

$\sqrt[3]{91125} = 3 \times 3 \times 5 = 45$

The cube root of 91125 is 45.

(i) Cube of any odd number is even.

False

Justification: The cube of an odd number is the product of the number with itself three times. The product of odd numbers is always odd. For example, $3^3 = 3 \times 3 \times 3 = 27$, which is an odd number.

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(ii) A perfect cube does not end with two zeros.

True

Justification: For a number to be a perfect cube, its prime factors must appear in groups of three. A number ending in zeros must be a multiple of 10, which has prime factors 2 and 5. For its cube to end in zeros, the number of zeros at the end must be a multiple of 3. For example, $10^3 = 1000$ (three zeros), $20^3 = 8000$ (three zeros), $100^3 = 1,000,000$ (six zeros). A number cannot have exactly two zeros at the end and be a perfect cube.

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(iii) If square of a number ends with 5, then its cube ends with 25.

False

Justification: If the square of a number ends with 5, the number itself must end with 5. Let's take the number 15. The square of 15 is $15^2 = 225$, which ends with 5. However, its cube is $15^3 = 3375$, which ends with 75, not 25.

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(iv) There is no perfect cube which ends with 8.

False

Justification: There are perfect cubes that end with the digit 8. For example, the cube of 2 is $2^3 = 8$. Another example is the cube of 12, which is $12^3 = 1728$.

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(v) The cube of a two digit number may be a three digit number.

False

Justification: The smallest two-digit number is 10. Its cube is $10^3 = 1000$, which is a four-digit number. The cube of any larger two-digit number will be even larger and will have at least four digits.

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(vi) The cube of a two digit number may have seven or more digits.

False

Justification: The largest two-digit number is 99. Let's estimate its cube. $99^3$ is slightly less than $100^3$. $100^3 = 1,000,000$, which has seven digits. The actual value of $99^3$ is $970,299$, which is a six-digit number. Therefore, the cube of a two-digit number can have at most six digits.

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(vii) The cube of a single digit number may be a single digit number.

True

Justification: The statement says it *may be* a single-digit number. This is true for some cases. For example, $1^3 = 1$ and $2^3 = 8$. Both 1 and 8 are single-digit numbers.

We can guess the cube root of a perfect cube by using the estimation method (grouping digits and looking at the unit and tens digits).

Let's recall the unit digits of cubes of numbers 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)

Notice the mapping of unit digits: $1 \to 1$, $2 \to 8$, $3 \to 7$, $4 \to 4$, $5 \to 5$, $6 \to 6$, $7 \to 3$, $8 \to 2$, $9 \to 9$, $0 \to 0$. The unit digits of the cube root are unique for each unit digit of the cube, except for the pairs (2,8) and (3,7) which swap unit digits.

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Guessing the cube root of 1331:

Group the digits from the right: 1 and 331.

The first group is 331. Its unit digit is 1. The unit digit of the cube root must be 1 (since $1^3$ ends in 1).

The second group is 1. We need to find a perfect cube less than or equal to 1. $1^3 = 1$. So, the tens digit of the cube root is 1.

Combining the tens digit (1) and the unit digit (1), the cube root is 11.

Check: $11^3 = 11 \times 11 \times 11 = 121 \times 11 = 1331$.

The cube root of 1331 is 11.

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Guessing the cube root of 4913:

Group the digits from the right: 4 and 913.

The first group is 913. Its unit digit is 3. The unit digit of the cube root must be 7 (since $7^3$ ends in 3).

The second group is 4. We need to find a perfect cube less than or equal to 4. $1^3 = 1$, $2^3 = 8$. $1 \leq 4 < 8$, so $1^3 \leq 4 < 2^3$. The smaller number is 1. The tens digit of the cube root is 1.

Combining the tens digit (1) and the unit digit (7), the cube root is 17.

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

The cube root of 4913 is 17.

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Guessing the cube root of 12167:

Group the digits from the right: 12 and 167.

The first group is 167. Its unit digit is 7. The unit digit of the cube root must be 3 (since $3^3$ ends in 7).

The second group is 12. We need to find a perfect cube less than or equal to 12. $2^3 = 8$, $3^3 = 27$. $8 \leq 12 < 27$, so $2^3 \leq 12 < 3^3$. The smaller number is 2. The tens digit of the cube root is 2.

Combining the tens digit (2) and the unit digit (3), the cube root is 23.

Check: $23^3 = 23 \times 23 \times 23 = 529 \times 23 = 12167$.

The cube root of 12167 is 23.

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Guessing the cube root of 32768:

Group the digits from the right: 32 and 768.

The first group is 768. Its unit digit is 8. The unit digit of the cube root must be 2 (since $2^3$ ends in 8).

The second group is 32. We need to find a perfect cube less than or equal to 32. $3^3 = 27$, $4^3 = 64$. $27 \leq 32 < 64$, so $3^3 \leq 32 < 4^3$. The smaller number is 3. The tens digit of the cube root is 3.

Combining the tens digit (3) and the unit digit (2), the cube root is 32.

Check: $32^3 = 32 \times 32 \times 32 = 1024 \times 32 = 32768$.

The cube root of 32768 is 32.