Chapter 9 Algebraic Expressions and Identities

Solution:

We need to add the three algebraic expressions:

1. $7xy + 5yz – 3zx$

2. $4yz + 9zx – 4y$

3. $–3xz + 5x – 2xy$ (Note that $xz$ is the same as $zx$)

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Method 1: Horizontal Addition

We write all the expressions in a single line and then group the like terms together.

$(7xy + 5yz – 3zx) + (4yz + 9zx – 4y) + (–2xy – 3zx + 5x)$

Removing the brackets:

$7xy + 5yz – 3zx + 4yz + 9zx – 4y – 2xy – 3zx + 5x$

Now, we rearrange and group the like terms:

$(7xy – 2xy) + (5yz + 4yz) + (–3zx + 9zx – 3zx) – 4y + 5x$

Combining the coefficients of the like terms:

$(7-2)xy + (5+4)yz + (-3+9-3)zx - 4y + 5x$

$5xy + 9yz + 3zx - 4y + 5x$

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Method 2: Columnar Addition

We arrange the expressions in rows with like terms vertically aligned in columns.

$\begin{array}{ccccccccc} & 7xy & + & 5yz & - & 3zx & & & \\ & & & 4yz & + & 9zx & - & 4y & \\ + & -2xy & & & - & 3zx & & & + 5x \\ \hline & 5xy & + & 9yz & + & 3zx & - & 4y & + 5x \\ \hline \end{array}$

Adding the coefficients in each column gives us the same result.

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Final Answer:

The sum of the given expressions is $5xy + 9yz + 3zx + 5x - 4y$.

Solution:

We need to subtract the expression $(5x^2 – 4y^2 + 6y – 3)$ from $(7x^2 – 4xy + 8y^2 + 5x – 3y)$.

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Method 1: Horizontal Subtraction

We write the expression as:

$(7x^2 – 4xy + 8y^2 + 5x – 3y) - (5x^2 – 4y^2 + 6y – 3)$

When we remove the second bracket, we change the sign of each term inside it:

$7x^2 – 4xy + 8y^2 + 5x – 3y - 5x^2 + 4y^2 - 6y + 3$

Now, we rearrange and group the like terms:

$(7x^2 - 5x^2) + (8y^2 + 4y^2) – 4xy + 5x + (-3y - 6y) + 3$

Combining the coefficients of the like terms:

$(7-5)x^2 + (8+4)y^2 - 4xy + 5x + (-3-6)y + 3$

$2x^2 + 12y^2 - 4xy + 5x - 9y + 3$

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Method 2: Columnar Subtraction

We write the expressions in two rows, aligning the like terms. Then we change the sign of each term in the bottom row (the expression being subtracted) and add the columns.

$\begin{array}{cccccccccc} & 7x^2 & + & 8y^2 & - & 4xy & + & 5x & - & 3y & \\ & 5x^2 & - & 4y^2 & & & & & + & 6y & - & 3 \\ &(-)& (+) & & & & & & (-)& & (+) \\ \hline \end{array}$

After changing the signs and adding:

$\begin{array}{cccccccccccc} & 7x^2 & + & 8y^2 & - & 4xy & + & 5x & - & 3y & \\ - & 5x^2 & + & 4y^2 & & & & & - & 6y & + & 3 \\ \hline & 2x^2 & + & 12y^2 & - & 4xy & + & 5x & - & 9y & + & 3 \\ \hline \end{array}$

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Final Answer:

The result of the subtraction is $2x^2 + 12y^2 - 4xy + 5x - 9y + 3$.

In an algebraic expression, terms are parts of the expression separated by addition or subtraction signs. The coefficient of a term is the numerical factor multiplying the variable part.

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(i) 5xyz² – 3zy:

• Term 1: $5xyz^2$. Coefficient: 5.
• Term 2: $-3zy$. Coefficient: -3.

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(ii) 1 + x + x²:

• Term 1: 1. Coefficient: 1.
• Term 2: x. Coefficient: 1.
• Term 3: $x^2$. Coefficient: 1.

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(iii) 4x²y² – 4x²y²z² + z²:

• Term 1: $4x^2y^2$. Coefficient: 4.
• Term 2: $-4x^2y^2z^2$. Coefficient: -4.
• Term 3: $z^2$. Coefficient: 1.

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(iv) 3 – pq + qr – rp:

• Term 1: 3. Coefficient: 3.
• Term 2: $-pq$. Coefficient: -1.
• Term 3: $+qr$. Coefficient: 1.
• Term 4: $-rp$. Coefficient: -1.

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(v) $\frac{x}{2}$ + $\frac{y}{2}$ − xy:

We can rewrite the terms as $\frac{1}{2}x$, $\frac{1}{2}y$, and $-1xy$.

• Term 1: $\frac{x}{2}$ or $\frac{1}{2}x$. Coefficient: $\frac{1}{2}$.
• Term 2: $\frac{y}{2}$ or $\frac{1}{2}y$. Coefficient: $\frac{1}{2}$.
• Term 3: $-xy$ or $-1xy$. Coefficient: -1.

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(vi) 0.3a – 0.6ab + 0.5b:

• Term 1: $0.3a$. Coefficient: 0.3.
• Term 2: $-0.6ab$. Coefficient: -0.6.
• Term 3: $0.5b$. Coefficient: 0.5.

A monomial is a polynomial with one term.

A binomial is a polynomial with two terms.

A trinomial is a polynomial with three terms.

Polynomials with more than three terms do not fit into these three categories.

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Let's classify each polynomial based on the number of terms:

• x + y : Has 2 terms (x and y). It is a Binomial.
• 1000 : Has 1 term (1000). It is a Monomial.
• x + x² + x³ + x⁴ : Has 4 terms (x, x², x³, and x⁴). It does not fit in any of the three categories (monomial, binomial, trinomial).
• 7 + y + 5x : Has 3 terms (7, y, and 5x). It is a Trinomial.
• 2y – 3y² : Has 2 terms (2y and $-3y^2$). It is a Binomial.
• 2y – 3y² + 4y³ : Has 3 terms (2y, $-3y^2$, and $4y^3$). It is a Trinomial.
• 5x – 4y + 3xy : Has 3 terms (5x, $-4y$, and 3xy). It is a Trinomial.
• 4z – 15z² : Has 2 terms (4z and $-15z^2$). It is a Binomial.
• ab + bc + cd + da : Has 4 terms (ab, bc, cd, and da). It does not fit in any of the three categories (monomial, binomial, trinomial).
• pqr : Has 1 term (pqr). It is a Monomial.
• p²q + pq² : Has 2 terms ($p^2q$ and $pq^2$). It is a Binomial.
• 2p + 2q : Has 2 terms (2p and 2q). It is a Binomial.

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Classification:

Monomials:

• 1000
• pqr

Binomials:

• x + y
• 2y – 3y²
• 4z – 15z²
• p²q + pq²
• 2p + 2q

Trinomials:

• 7 + y + 5x
• 2y – 3y² + 4y³
• 5x – 4y + 3xy

Polynomials that do not fit in any of these three categories:

• x + x² + x³ + x⁴
• ab + bc + cd + da

To add algebraic expressions, we combine like terms. We can either arrange them vertically or add them horizontally by grouping like terms.

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(i) ab – bc, bc – ca, ca – ab

Arrange vertically:

$\begin{array}{rccc} & ab & -bc & \\ & & +bc & -ca \\ & -ab & & +ca \\ \hline \end{array}$

Add the coefficients in each column:

• ab terms: $1ab - 1ab = (1-1)ab = 0ab = 0$
• bc terms: $-1bc + 1bc = (-1+1)bc = 0bc = 0$
• ca terms: $-1ca + 1ca = (-1+1)ca = 0ca = 0$

Sum = $0 + 0 + 0 = 0$.

Alternatively, horizontally:

Sum = 0.

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(ii) a – b + ab, b – c + bc, c – a + ac

Arrange vertically:

$\begin{array}{rccccc} & a & -b & +ab & & \\ & & +b & & -c & +bc \\ & -a & & & +c & & +ac \\ \hline \end{array}$

Add the coefficients in each column:

• a terms: $1a - 1a = (1-1)a = 0$
• b terms: $-1b + 1b = (-1+1)b = 0$
• ab terms: $+ab$ (no other ab terms)
• c terms: $-c + c = (-1+1)c = 0$
• bc terms: $+bc$ (no other bc terms)
• ac terms: $+ac$ (no other ac terms)

Sum = $0 + 0 + ab + 0 + bc + ac = ab + bc + ac$.

Alternatively, horizontally:

Sum = ab + bc + ac.

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(iii) 2p²q² – 3pq + 4, 5 + 7pq – 3p²q²

Arrange vertically (aligning like terms):

$\begin{array}{rccc} & 2p^2q^2 & -3pq & +4 \\ & -3p^2q^2 & +7pq & +5 \\ \hline \end{array}$

Add the coefficients in each column:

• $p^2q^2$ terms: $2p^2q^2 - 3p^2q^2 = (2-3)p^2q^2 = -1p^2q^2 = -p^2q^2$
• pq terms: $-3pq + 7pq = (-3+7)pq = 4pq$
• Constant terms: $+4 + 5 = 9$

Sum = $-p^2q^2 + 4pq + 9$.

Alternatively, horizontally:

Sum = $-p^2q^2 + 4pq + 9$.

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(iv) l² + m², m² + n², n² + l², 2lm + 2mn + 2nl

Arrange vertically:

$\begin{array}{rccccc} & l^2 & +m^2 & & & & \\ & & +m^2 & +n^2 & & & \\ & +l^2 & & +n^2 & & & \\ & & & & +2lm & +2mn & +2nl \\ \hline \end{array}$

Add the coefficients in each column:

• $l^2$ terms: $1l^2 + 1l^2 = (1+1)l^2 = 2l^2$
• $m^2$ terms: $1m^2 + 1m^2 = (1+1)m^2 = 2m^2$
• $n^2$ terms: $1n^2 + 1n^2 = (1+1)n^2 = 2n^2$
• lm terms: $+2lm$ (no other lm terms)
• mn terms: $+2mn$ (no other mn terms)
• nl terms: $+2nl$ (no other nl terms)

Sum = $2l^2 + 2m^2 + 2n^2 + 2lm + 2mn + 2nl$.

Alternatively, horizontally:

This expression can also be written as $2(l^2 + m^2 + n^2 + lm + mn + nl)$.

Sum = $2l^2 + 2m^2 + 2n^2 + 2lm + 2mn + 2nl$.

To subtract one algebraic expression from another, we add the additive inverse of the expression being subtracted to the other expression. To find the additive inverse, we change the sign of each term in the expression being subtracted.

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(a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3

Expression to be subtracted: $4a – 7ab + 3b + 12$

Additive inverse: $-(4a – 7ab + 3b + 12) = -4a + 7ab - 3b - 12$

Expression from which we subtract: $12a – 9ab + 5b – 3$

Add the second expression and the additive inverse of the first expression vertically, aligning like terms:

$\begin{array}{rcccc} & 12a & -9ab & +5b & -3 \\ + & -4a & +7ab & -3b & -12 \\ \hline \end{array}$

Add the coefficients in each column:

• a terms: $12a - 4a = (12-4)a = 8a$
• ab terms: $-9ab + 7ab = (-9+7)ab = -2ab$
• b terms: $+5b - 3b = (5-3)b = 2b$
• Constant terms: $-3 - 12 = -15$

Result = $8a - 2ab + 2b - 15$.

Alternatively, horizontally:

Result = $8a - 2ab + 2b - 15$.

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(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz

Expression to be subtracted: $3xy + 5yz – 7zx$

Additive inverse: $-(3xy + 5yz – 7zx) = -3xy - 5yz + 7zx$

Expression from which we subtract: $5xy – 2yz – 2zx + 10xyz$

Add the second expression and the additive inverse of the first expression vertically, aligning like terms:

$\begin{array}{rccccc} & 5xy & -2yz & -2zx & +10xyz \\ + & -3xy & -5yz & +7zx & & \\ \hline \end{array}$

Add the coefficients in each column:

• xy terms: $5xy - 3xy = (5-3)xy = 2xy$
• yz terms: $-2yz - 5yz = (-2-5)yz = -7yz$
• zx terms: $-2zx + 7zx = (-2+7)zx = 5zx$
• xyz terms: $+10xyz$ (no other xyz terms)

Result = $2xy - 7yz + 5zx + 10xyz$.

Alternatively, horizontally:

Result = $2xy - 7yz + 5zx + 10xyz$.

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(c) Subtract 4p²q – 3pq + 5pq² – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq² + 5p²q

Expression to be subtracted: $4p^2q – 3pq + 5pq^2 – 8p + 7q – 10$

Additive inverse: $-(4p^2q – 3pq + 5pq^2 – 8p + 7q – 10) = -4p^2q + 3pq - 5pq^2 + 8p - 7q + 10$

Expression from which we subtract: $18 – 3p – 11q + 5pq – 2pq^2 + 5p^2q$

Arrange the terms vertically, aligning like terms, and then add them:

$\begin{array}{rcccccccc} & 5p^2q & -2pq^2 & +5pq & -3p & -11q & +18 \\ + & -4p^2q & -5pq^2 & +3pq & +8p & -7q & +10 \\ \hline \end{array}$

Add the coefficients in each column:

• $p^2q$ terms: $5p^2q - 4p^2q = (5-4)p^2q = 1p^2q = p^2q$
• $pq^2$ terms: $-2pq^2 - 5pq^2 = (-2-5)pq^2 = -7pq^2$
• pq terms: $+5pq + 3pq = (5+3)pq = 8pq$
• p terms: $-3p + 8p = (-3+8)p = 5p$
• q terms: $-11q - 7q = (-11-7)q = -18q$
• Constant terms: $+18 + 10 = 28$

Result = $p^2q - 7pq^2 + 8pq + 5p - 18q + 28$.

Alternatively, horizontally:

Group like terms together:

Result = $p^2q - 7pq^2 + 8pq + 5p - 18q + 28$.

Solution:

The area of a rectangle is calculated by the formula:

Area = Length × Breadth

We will calculate the area for each row by multiplying the given monomials for length and breadth.

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For the second row:

Length = $9y$

Breadth = $4y^2$

Area = $(9y) \times (4y^2)$

$\implies (9 \times 4) \times (y \times y^2) = 36 \times y^{1+2} = 36y^3$

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For the third row:

Length = $4ab$

Breadth = $5bc$

Area = $(4ab) \times (5bc)$

$\implies (4 \times 5) \times (a \times b \times b \times c) = 20 \times ab^2c = 20ab^2c$

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For the fourth row:

Length = $2l^2m$

Breadth = $3lm^2$

Area = $(2l^2m) \times (3lm^2)$

$\implies (2 \times 3) \times (l^2 \times l \times m \times m^2) = 6 \times l^{2+1}m^{1+2} = 6l^3m^3$

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The completed table is as follows:

Length	Breadth	Area
$3x$	$5y$	$15xy$
$9y$	$4y^2$	$36y^3$
$4ab$	$5bc$	$20ab^2c$
$2l^2m$	$3lm^2$	$6l^3m^3$

Solution:

The volume of a rectangular box (cuboid) is given by the formula:

Volume = Length × Breadth × Height

We will find the volume for each case by multiplying the three given monomials.

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(i) Length = $2ax$, Breadth = $3by$, Height = $5cz$

Volume = $(2ax) \times (3by) \times (5cz)$

$\implies (2 \times 3 \times 5) \times (a \times x \times b \times y \times c \times z)$

$\implies 30 \times (abcxyz)$

$\implies 30abcxyz$

The volume of the box is $30abcxyz$.

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(ii) Length = $m^2n$, Breadth = $n^2p$, Height = $p^2m$

Volume = $(m^2n) \times (n^2p) \times (p^2m)$

$\implies (m^2 \times m) \times (n \times n^2) \times (p \times p^2)$

$\implies m^{2+1} \times n^{1+2} \times p^{1+2}$

$\implies m^3n^3p^3$

The volume of the box is $m^3n^3p^3$.

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(iii) Length = $2q$, Breadth = $4q^2$, Height = $8q^3$

Volume = $(2q) \times (4q^2) \times (8q^3)$

$\implies (2 \times 4 \times 8) \times (q \times q^2 \times q^3)$

$\implies 64 \times q^{1+2+3}$

$\implies 64q^6$

The volume of the box is $64q^6$.

To find the product of monomials, multiply the coefficients together and multiply the variable parts together. Remember the exponent rule $a^m \times a^n = a^{m+n}$.

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(i) 4, 7p:

Product = $28p$.

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(ii) – 4p, 7p:

Product = $-28p^2$.

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(iii) – 4p, 7pq:

Product = $-28p^2q$.

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(iv) 4p³, – 3p:

Product = $-12p^4$.

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(v) 4p, 0:

Product = 0.

The area of a rectangle is given by the formula: Area = Length $\times$ Breadth.

We need to multiply the given pairs of monomials.

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(p, q):

Length = p, Breadth = q

Area = $pq$.

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(10m, 5n):

Length = 10m, Breadth = 5n

Area = $50mn$.

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(20x², 5y²):

Length = $20x^2$, Breadth = $5y^2$

Area = $100x^2y^2$.

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(4x, 3x²):

Length = 4x, Breadth = $3x^2$

Area = $12x^3$.

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(3mn, 4np):

Length = 3mn, Breadth = 4np

Area = $12mn^2p$.

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Here is a summary of the areas:

• For (p, q): Area = $pq$
• For (10m, 5n): Area = $50mn$
• For (20x², 5y²): Area = $100x^2y^2$
• For (4x, 3x²): Area = $12x^3$
• For (3mn, 4np): Area = $12mn^2p$

To complete the table, we need to find the product of the monomial in the first column (Second Monomial) and the monomial in the first row (First Monomial). The product goes into the cell at their intersection.

Remember to multiply the coefficients and combine the variable terms using the rule $a^m \times a^n = a^{m+n}$.

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Here is the completed table of products:

First Monomial $\rightarrow$ Second Monomial $\downarrow$	$2x$	$-5y$	$3x^2$	$-4xy$	$7x^2y$	$-9x^2y^2$
$2x$	$4x^2$	$-10xy$	$6x^3$	$-8x^2y$	$14x^3y$	$-18x^3y^2$
$-5y$	$-10xy$	$25y^2$	$-15x^2y$	$20xy^2$	$-35x^2y^2$	$45x^2y^3$
$3x^2$	$6x^3$	$-15x^2y$	$9x^4$	$-12x^3y$	$21x^4y$	$-27x^4y^2$
$-4xy$	$-8x^2y$	$20xy^2$	$-12x^3y$	$16x^2y^2$	$-28x^3y^2$	$36x^3y^3$
$7x^2y$	$14x^3y$	$-35x^2y^2$	$21x^4y$	$-28x^3y^2$	$49x^4y^2$	$-63x^4y^3$
$-9x^2y^2$	$-18x^3y^2$	$45x^2y^3$	$-27x^4y^2$	$36x^3y^3$	$-63x^4y^3$	$81x^4y^4$

Solution:

The volume of a rectangular box is given by the formula:

Volume = Length × Breadth × Height

We will calculate the volume for each case by multiplying the given expressions for length, breadth, and height.

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(i) Length = $5a$, Breadth = $3a^2$, Height = $7a^4$

Volume = $(5a) \times (3a^2) \times (7a^4)$

First, we multiply the coefficients:

$5 \times 3 \times 7 = 105$

Next, we multiply the variables:

$a \times a^2 \times a^4 = a^{1+2+4} = a^7$

Combining them, we get:

Volume = $105a^7$

The volume of the rectangular box is $105a^7$.

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(ii) Length = $2p$, Breadth = $4q$, Height = $8r$

Volume = $(2p) \times (4q) \times (8r)$

Multiplying the coefficients:

$2 \times 4 \times 8 = 64$

Multiplying the variables:

$p \times q \times r = pqr$

Combining them, we get:

Volume = $64pqr$

The volume of the rectangular box is $64pqr$.

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(iii) Length = $xy$, Breadth = $2x^2y$, Height = $2xy^2$

Volume = $(xy) \times (2x^2y) \times (2xy^2)$

Multiplying the coefficients:

$1 \times 2 \times 2 = 4$

Multiplying the variables (grouping like variables):

$(x \times x^2 \times x) \times (y \times y \times y^2) = (x^{1+2+1}) \times (y^{1+1+2}) = x^4y^4$

Combining them, we get:

Volume = $4x^4y^4$

The volume of the rectangular box is $4x^4y^4$.

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(iv) Length = $a$, Breadth = $2b$, Height = $3c$

Volume = $(a) \times (2b) \times (3c)$

Multiplying the coefficients:

$1 \times 2 \times 3 = 6$

Multiplying the variables:

$a \times b \times c = abc$

Combining them, we get:

Volume = $6abc$

The volume of the rectangular box is $6abc$.

To find the product of the given monomials, we multiply their coefficients and their variable parts separately.

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(i) xy, yz, zx

Product = $(xy) \times (yz) \times (zx)$

We rearrange the variables to group the like ones together:

Product = $(x \times x) \times (y \times y) \times (z \times z)$

Using the rule of exponents, $a^m \times a^n = a^{m+n}$:

Product = $x^{1+1} \times y^{1+1} \times z^{1+1}$

Product = $x^2y^2z^2$

The product is $x^2y^2z^2$.

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(ii) a, – a², a³

Product = $(a) \times (–a^2) \times (a^3)$

First, we determine the sign of the product. Since there is one negative term, the result will be negative.

Product = $-(a \times a^2 \times a^3)$

Now, we add the powers of the variable 'a':

Product = $- (a^{1+2+3})$

Product = $-a^6$

The product is $-a^6$.

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(iii) 2, 4y, 8y², 16y³

Product = $(2) \times (4y) \times (8y^2) \times (16y^3)$

First, we multiply the numerical coefficients:

$2 \times 4 \times 8 \times 16 = 8 \times 8 \times 16 = 64 \times 16 = 1024$

Next, we multiply the variable parts:

$y \times y^2 \times y^3 = y^{1+2+3} = y^6$

Combining the results:

Product = $1024y^6$

The product is $1024y^6$.

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(iv) a, 2b, 3c, 6abc

Product = $(a) \times (2b) \times (3c) \times (6abc)$

First, we multiply the coefficients:

$1 \times 2 \times 3 \times 6 = 36$

Next, we multiply the variables, grouping like variables together:

$(a \times a) \times (b \times b) \times (c \times c) = a^2 \times b^2 \times c^2 = a^2b^2c^2$

Combining the results:

Product = $36a^2b^2c^2$

The product is $36a^2b^2c^2$.

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(v) m, – mn, mnp

Product = $(m) \times (–mn) \times (mnp)$

First, we determine the sign. Since there is one negative term, the product is negative.

Product = $- (m \times mn \times mnp)$

Now, we group the like variables:

Product = $-((m \times m \times m) \times (n \times n) \times (p))$

Adding the powers of the variables:

Product = $-(m^{1+1+1} \times n^{1+1} \times p^1)$

Product = $-m^3n^2p$

The product is $-m^3n^2p$.

(i) x (x – 3) + 2 for x = 1

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Step 1: Simplify the expression

First, we distribute $x$ over the term $(x-3)$:

$x(x-3) + 2 = (x \times x) - (x \times 3) + 2$

$= x^2 - 3x + 2$

The simplified expression is $x^2 - 3x + 2$.

Step 2: Evaluate the simplified expression for x = 1

Now, we substitute $x = 1$ into the simplified expression:

$(1)^2 - 3(1) + 2$

$= 1 - 3 + 2$

$= 3 - 3 = 0$

The value of the expression is 0.

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(ii) 3y (2y – 7) – 3 (y – 4) – 63 for y = –2

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Step 1: Simplify the expression

We use the distributive property for both parts of the expression:

$3y(2y-7) - 3(y-4) - 63$

$= (3y \times 2y - 3y \times 7) - (3 \times y - 3 \times 4) - 63$

$= (6y^2 - 21y) - (3y - 12) - 63$

Now, we remove the brackets. Remember to change the signs of the terms in the second bracket.

$= 6y^2 - 21y - 3y + 12 - 63$

Group and combine the like terms:

$= 6y^2 + (-21y - 3y) + (12 - 63)$

$= 6y^2 - 24y - 51$

The simplified expression is $6y^2 - 24y - 51$.

Step 2: Evaluate the simplified expression for y = –2

Substitute $y = -2$ into the simplified expression:

$6(-2)^2 - 24(-2) - 51$

$= 6(4) - (-48) - 51$

$= 24 + 48 - 51$

$= 72 - 51 = 21$

The value of the expression is 21.

(i) Add 5m (3 – m) and 6m² – 13m

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Step 1: Simplify the first expression

We simplify $5m(3-m)$ by using the distributive property:

$5m(3-m) = (5m \times 3) - (5m \times m) = 15m - 5m^2$

Step 2: Add the simplified expression to the second expression

We need to add $(15m - 5m^2)$ and $(6m^2 - 13m)$.

$(15m - 5m^2) + (6m^2 - 13m)$

Group the like terms together:

$(-5m^2 + 6m^2) + (15m - 13m)$

Combine the like terms:

$m^2 + 2m$

The sum is $m^2 + 2m$.

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(ii) Add 4y (3y² + 5y – 7) and 2 (y³ – 4y² + 5)

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Step 1: Simplify both expressions

First expression: $4y(3y^2 + 5y - 7)$

$= (4y \times 3y^2) + (4y \times 5y) - (4y \times 7)$

$= 12y^3 + 20y^2 - 28y$

Second expression: $2(y^3 - 4y^2 + 5)$

$= (2 \times y^3) - (2 \times 4y^2) + (2 \times 5)$

$= 2y^3 - 8y^2 + 10$

Step 2: Add the two simplified expressions

We need to add $(12y^3 + 20y^2 - 28y)$ and $(2y^3 - 8y^2 + 10)$.

$(12y^3 + 20y^2 - 28y) + (2y^3 - 8y^2 + 10)$

Group the like terms together:

$(12y^3 + 2y^3) + (20y^2 - 8y^2) - 28y + 10$

Combine the like terms:

$14y^3 + 12y^2 - 28y + 10$

The sum is $14y^3 + 12y^2 - 28y + 10$.

Solution:

We need to calculate: $2pq(p+q) - 3pq(p-q)$

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Step 1: Simplify both expressions

First expression: $2pq(p+q)$

$= (2pq \times p) + (2pq \times q)$

$= 2p^2q + 2pq^2$

Second expression: $3pq(p-q)$

$= (3pq \times p) - (3pq \times q)$

$= 3p^2q - 3pq^2$

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Step 2: Perform the subtraction

We subtract the simplified second expression from the simplified first expression:

$(2p^2q + 2pq^2) - (3p^2q - 3pq^2)$

Remove the bracket, changing the sign of each term inside it:

$2p^2q + 2pq^2 - 3p^2q + 3pq^2$

Group the like terms together:

$(2p^2q - 3p^2q) + (2pq^2 + 3pq^2)$

Combine the like terms:

$-p^2q + 5pq^2$

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Final Answer:

The result of the subtraction is $-p^2q + 5pq^2$ or $5pq^2 - p^2q$.

To carry out the multiplication of a monomial with a polynomial, we use the distributive property, which states that $a(b+c) = ab + ac$. We multiply the monomial with each term of the polynomial.

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(i) 4p, q + r

We need to multiply the monomial $4p$ with the binomial $(q+r)$.

Product = $4p \times (q+r)$

Using the distributive property:

Product = $(4p \times q) + (4p \times r)$

Product = $4pq + 4pr$

The product is $4pq + 4pr$.

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(ii) ab, a – b

We need to multiply the monomial $ab$ with the binomial $(a-b)$.

Product = $ab \times (a-b)$

Using the distributive property:

Product = $(ab \times a) - (ab \times b)$

Product = $a^2b - ab^2$

The product is $a^2b - ab^2$.

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(iii) a + b, 7a²b²

We need to multiply the binomial $(a+b)$ with the monomial $7a^2b^2$.

Product = $(a+b) \times (7a^2b^2)$

Using the distributive property:

Product = $(a \times 7a^2b^2) + (b \times 7a^2b^2)$

Product = $7a^3b^2 + 7a^2b^3$

The product is $7a^3b^2 + 7a^2b^3$.

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(iv) a² – 9, 4a

We need to multiply the binomial $(a^2 - 9)$ with the monomial $4a$.

Product = $(a^2 - 9) \times (4a)$

Using the distributive property:

Product = $(a^2 \times 4a) - (9 \times 4a)$

Product = $4a^3 - 36a$

The product is $4a^3 - 36a$.

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(v) pq + qr + rp, 0

We need to multiply the trinomial $(pq + qr + rp)$ with the monomial $0$.

Product = $(pq + qr + rp) \times 0$

According to the zero property of multiplication, any expression multiplied by zero results in zero.

Product = $0$

The product is $0$.

To complete the table, we need to find the product of the first and second expressions for each row. This involves multiplying a monomial by a polynomial, for which we use the distributive property.

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(i) First expression: $a$, Second expression: $b + c + d$

Product = $a \times (b + c + d)$

Applying the distributive property:

Product = $(a \times b) + (a \times c) + (a \times d)$

Product = $ab + ac + ad$

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(ii) First expression: $x + y – 5$, Second expression: $5xy$

Product = $(x + y - 5) \times (5xy)$

Applying the distributive property:

Product = $(x \times 5xy) + (y \times 5xy) - (5 \times 5xy)$

Product = $5x^2y + 5xy^2 - 25xy$

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(iii) First expression: $p$, Second expression: $6p^2 – 7p + 5$

Product = $p \times (6p^2 - 7p + 5)$

Applying the distributive property:

Product = $(p \times 6p^2) - (p \times 7p) + (p \times 5)$

Product = $6p^3 - 7p^2 + 5p$

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(iv) First expression: $4p^2q^2$, Second expression: $p^2 – q^2$

Product = $4p^2q^2 \times (p^2 - q^2)$

Applying the distributive property:

Product = $(4p^2q^2 \times p^2) - (4p^2q^2 \times q^2)$

Product = $4p^4q^2 - 4p^2q^4$

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(v) First expression: $a + b + c$, Second expression: $abc$

Product = $(a + b + c) \times (abc)$

Applying the distributive property:

Product = $(a \times abc) + (b \times abc) + (c \times abc)$

Product = $a^2bc + ab^2c + abc^2$

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Completed Table:

S.No.	First expression	Second expression	Product
(i)	$a$	$b + c + d$	$ab + ac + ad$
(ii)	$x + y – 5$	$5xy$	$5x^2y + 5xy^2 - 25xy$
(iii)	$p$	$6p^2 – 7p + 5$	$6p^3 - 7p^2 + 5p$
(iv)	$4p^2q^2$	$p^2 – q^2$	$4p^4q^2 - 4p^2q^4$
(v)	$a + b + c$	$abc$	$a^2bc + ab^2c + abc^2$

To find the product of monomials, we multiply their numerical coefficients and then multiply their variable parts. When multiplying variables with the same base, we add their exponents ($a^m \times a^n = a^{m+n}$).

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(i) (a²) × (2a²²) × (4a²⁶)

Product = $(1 \times 2 \times 4) \times (a^2 \times a^{22} \times a^{26})$

Multiply the coefficients:

$1 \times 2 \times 4 = 8$

Multiply the variable parts by adding the exponents:

$a^{2 + 22 + 26} = a^{50}$

So, the product is $8a^{50}$.

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(ii) $\left( \frac{2}{3}xy \right)\times\left( \frac{-9}{10}x^2y^2 \right)$

Product = $\left( \frac{2}{3} \times \frac{-9}{10} \right) \times (x \times x^2) \times (y \times y^2)$

Multiply the coefficients:

$\frac{2}{3} \times \frac{-9}{10} = \frac{\cancel{2}^1}{\cancel{3}_1} \times \frac{\cancel{-9}^{-3}}{\cancel{10}_5} = \frac{1 \times -3}{1 \times 5} = -\frac{3}{5}$

Multiply the variable parts by adding the exponents:

$(x^{1+2}) \times (y^{1+2}) = x^3y^3$

So, the product is $-\frac{3}{5}x^3y^3$.

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(iii) $\left( -\frac{10}{3}pq^3 \right)\times\left( \frac{6}{5}p^3q \right)$

Product = $\left( -\frac{10}{3} \times \frac{6}{5} \right) \times (p \times p^3) \times (q^3 \times q)$

Multiply the coefficients:

$-\frac{10}{3} \times \frac{6}{5} = -\frac{\cancel{10}^2}{\cancel{3}_1} \times \frac{\cancel{6}^2}{\cancel{5}_1} = -(2 \times 2) = -4$

Multiply the variable parts by adding the exponents:

$(p^{1+3}) \times (q^{3+1}) = p^4q^4$

So, the product is $-4p^4q^4$.

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(iv) x × x² × x³ × x⁴

Product = $x^1 \times x^2 \times x^3 \times x^4$

The coefficient is 1. We multiply the variable parts by adding the exponents:

Product = $x^{1+2+3+4}$

Product = $x^{10}$

So, the product is $x^{10}$.

(a) Simplify 3x (4x – 5) + 3 and find its values

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Step 1: Simplify the expression

We use the distributive property to multiply $3x$ with each term inside the bracket $(4x - 5)$.

$3x (4x – 5) + 3 = (3x \times 4x) - (3x \times 5) + 3$

$= 12x^2 - 15x + 3$

The simplified expression is $12x^2 - 15x + 3$.

Step 2: Find the values for the given values of x

(i) For x = 3

We substitute $x = 3$ into the simplified expression:

$12(3)^2 - 15(3) + 3$

$= 12(9) - 45 + 3$

$= 108 - 45 + 3$

$= 63 + 3 = 66$

The value of the expression is 66.

(ii) For x = $\frac{1}{2}$

We substitute $x = \frac{1}{2}$ into the simplified expression:

$12\left(\frac{1}{2}\right)^2 - 15\left(\frac{1}{2}\right) + 3$

$= 12\left(\frac{1}{4}\right) - \frac{15}{2} + 3$

$= \frac{12}{4} - \frac{15}{2} + 3$

$= 3 - \frac{15}{2} + 3$

$= 6 - \frac{15}{2}$

$= \frac{12}{2} - \frac{15}{2} = \frac{12-15}{2} = -\frac{3}{2}$

The value of the expression is $-\frac{3}{2}$.

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(b) Simplify a (a² + a + 1) + 5 and find its value

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Step 1: Simplify the expression

We use the distributive property to multiply $a$ with each term inside the bracket $(a^2 + a + 1)$.

$a (a^2 + a + 1) + 5 = (a \times a^2) + (a \times a) + (a \times 1) + 5$

$= a^3 + a^2 + a + 5$

The simplified expression is $a^3 + a^2 + a + 5$.

Step 2: Find the values for the given values of a

(i) For a = 0

Substitute $a=0$ into the simplified expression:

$(0)^3 + (0)^2 + (0) + 5$

$= 0 + 0 + 0 + 5 = 5$

The value of the expression is 5.

(ii) For a = 1

Substitute $a=1$ into the simplified expression:

$(1)^3 + (1)^2 + (1) + 5$

$= 1 + 1 + 1 + 5 = 8$

The value of the expression is 8.

(iii) For a = – 1

Substitute $a=-1$ into the simplified expression:

$(-1)^3 + (-1)^2 + (-1) + 5$

$= -1 + 1 - 1 + 5$

$= 0 - 1 + 5 = 4$

The value of the expression is 4.

(a) Add: p ( p – q), q ( q – r) and r ( r – p)

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Step 1: Simplify each expression using the distributive property.

First expression: $p(p-q) = p \times p - p \times q = p^2 - pq$

Second expression: $q(q-r) = q \times q - q \times r = q^2 - qr$

Third expression: $r(r-p) = r \times r - r \times p = r^2 - rp$

Step 2: Add the simplified expressions.

We need to find the sum: $(p^2 - pq) + (q^2 - qr) + (r^2 - rp)$

Removing the brackets and rearranging the terms:

$p^2 + q^2 + r^2 - pq - qr - rp$

Since there are no like terms to combine, this is the final answer.

The sum is $p^2 + q^2 + r^2 - pq - qr - rp$.

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(b) Add: 2x (z – x – y) and 2y (z – y – x)

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Step 1: Simplify each expression.

First expression: $2x(z - x - y) = (2x \times z) - (2x \times x) - (2x \times y) \ $$ = 2xz - 2x^2 - 2xy$

Second expression: $2y(z - y - x) = (2y \times z) - (2y \times y) - (2y \times x) \ $$ = 2yz - 2y^2 - 2xy$

Step 2: Add the simplified expressions.

$(2xz - 2x^2 - 2xy) + (2yz - 2y^2 - 2xy)$

Group the like terms together:

$-2x^2 - 2y^2 + (-2xy - 2xy) + 2yz + 2xz$

Combine the like terms:

$-2x^2 - 2y^2 - 4xy + 2yz + 2xz$

The sum is $-2x^2 - 2y^2 - 4xy + 2yz + 2xz$.

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(c) Subtract: 3l (l – 4m + 5n) from 4l (10n – 3m + 2l)

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Step 1: Simplify both expressions.

Expression to be subtracted: $3l(l - 4m + 5n) = 3l^2 - 12lm + 15ln$

Expression to subtract from: $4l(10n - 3m + 2l) = 40ln - 12lm + 8l^2$

Step 2: Perform the subtraction.

$(40ln - 12lm + 8l^2) - (3l^2 - 12lm + 15ln)$

Remove the brackets, changing the signs of the second expression:

$40ln - 12lm + 8l^2 - 3l^2 + 12lm - 15ln$

Group and combine like terms:

$(8l^2 - 3l^2) + (-12lm + 12lm) + (40ln - 15ln)$

$5l^2 + 0 + 25ln$

$5l^2 + 25ln$

The result is $5l^2 + 25ln$.

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(d) Subtract: 3a (a + b + c) – 2b (a – b + c) from 4c (– a + b + c )

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Step 1: Simplify both expressions.

Expression to be subtracted:

$3a(a+b+c) - 2b(a-b+c)$

$= (3a^2 + 3ab + 3ac) - (2ab - 2b^2 + 2bc)$

$= 3a^2 + 3ab + 3ac - 2ab + 2b^2 - 2bc$

$= 3a^2 + (3ab - 2ab) + 2b^2 + 3ac - 2bc$

$= 3a^2 + ab + 2b^2 + 3ac - 2bc$

Expression to subtract from:

$4c(-a + b + c) = -4ac + 4bc + 4c^2$

Step 2: Perform the subtraction.

$(-4ac + 4bc + 4c^2) - (3a^2 + ab + 2b^2 + 3ac - 2bc)$

Remove the brackets, changing the signs of the second expression:

$-4ac + 4bc + 4c^2 - 3a^2 - ab - 2b^2 - 3ac + 2bc$

Group and combine like terms:

$-3a^2 - 2b^2 + 4c^2 - ab + (4bc + 2bc) + (-4ac - 3ac)$

$-3a^2 - 2b^2 + 4c^2 - ab + 6bc - 7ac$

The result is $-3a^2 - 2b^2 + 4c^2 - ab + 6bc - 7ac$.

To multiply two binomials, we can use the distributive property. Each term of the first binomial is multiplied by each term of the second binomial. This is often remembered by the acronym FOIL (First, Outer, Inner, Last).

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(i) (x – 4) and (2x + 3)

Product = $(x - 4)(2x + 3)$

Using the distributive property:

$= x(2x+3) - 4(2x+3)$

$= (x \times 2x) + (x \times 3) - (4 \times 2x) - (4 \times 3)$

$= 2x^2 + 3x - 8x - 12$

Now, combine the like terms (the middle terms):

$= 2x^2 + (3-8)x - 12$

$= 2x^2 - 5x - 12$

The product is $2x^2 - 5x - 12$.

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(ii) (x – y) and (3x + 5y)

Product = $(x - y)(3x + 5y)$

Using the distributive property:

$= x(3x+5y) - y(3x+5y)$

$= (x \times 3x) + (x \times 5y) - (y \times 3x) - (y \times 5y)$

$= 3x^2 + 5xy - 3xy - 5y^2$

Now, combine the like terms (the middle terms):

$= 3x^2 + (5-3)xy - 5y^2$

$= 3x^2 + 2xy - 5y^2$

The product is $3x^2 + 2xy - 5y^2$.

We use the distributive property to multiply the two expressions. Each term of the first expression is multiplied by each term of the second expression.

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(i) (a + 7) and (b – 5)

Product = $(a + 7)(b - 5)$

$= a(b - 5) + 7(b - 5)$

$= (a \times b) - (a \times 5) + (7 \times b) - (7 \times 5)$

$= ab - 5a + 7b - 35$

Since there are no like terms to combine, this is the final simplified expression.

The product is $ab - 5a + 7b - 35$.

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(ii) (a² + 2b²) and (5a – 3b)

Product = $(a^2 + 2b^2)(5a - 3b)$

$= a^2(5a - 3b) + 2b^2(5a - 3b)$

$= (a^2 \times 5a) - (a^2 \times 3b) + (2b^2 \times 5a) - (2b^2 \times 3b)$

$= 5a^3 - 3a^2b + 10ab^2 - 6b^3$

There are no like terms to combine in this result.

The product is $5a^3 - 3a^2b + 10ab^2 - 6b^3$.

Solution:

We need to simplify the expression $(a + b) (2a – 3b + c) – (2a – 3b) c$.

We will handle the two parts of the expression separately and then combine them.

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Part 1: Simplify $(a + b) (2a – 3b + c)$

We multiply each term of $(a+b)$ by each term of $(2a - 3b + c)$.

$a(2a - 3b + c) + b(2a - 3b + c)$

$= (a \times 2a) - (a \times 3b) + (a \times c) + (b \times 2a) - (b \times 3b) + (b \times c)$

$= 2a^2 - 3ab + ac + 2ab - 3b^2 + bc$

Combine like terms ($-3ab$ and $+2ab$):

$= 2a^2 - ab + ac - 3b^2 + bc$

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Part 2: Simplify $– (2a – 3b) c$

We multiply $-c$ with each term of $(2a - 3b)$.

$-c(2a - 3b) = -(c \times 2a) - (-c \times 3b)$

$= -2ac + 3bc$

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Step 3: Combine the results from Part 1 and Part 2

Now we add the simplified expressions:

$(2a^2 - ab + ac - 3b^2 + bc) + (-2ac + 3bc)$

$= 2a^2 - ab + ac - 3b^2 + bc - 2ac + 3bc$

Group and combine the like terms:

$= 2a^2 - 3b^2 - ab + (bc + 3bc) + (ac - 2ac)$

$= 2a^2 - 3b^2 - ab + 4bc - ac$

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Final Answer:

The simplified expression is $2a^2 - 3b^2 - ab + 4bc - ac$.

To multiply the binomials, we use the distributive property, where each term of the first binomial is multiplied by each term of the second binomial. The general form is $(a+b)(c+d) = a(c+d) + b(c+d) = ac + ad + bc + bd$.

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(i) (2x + 5) and (4x – 3)

Product = $(2x + 5)(4x - 3)$

$= 2x(4x - 3) + 5(4x - 3)$

$= (2x \times 4x) - (2x \times 3) + (5 \times 4x) - (5 \times 3)$

$= 8x^2 - 6x + 20x - 15$

Combine the like terms ($-6x$ and $+20x$):

$= 8x^2 + 14x - 15$

The product is $8x^2 + 14x - 15$.

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(ii) (y – 8) and (3y – 4)

Product = $(y - 8)(3y - 4)$

$= y(3y - 4) - 8(3y - 4)$

$= (y \times 3y) - (y \times 4) - (8 \times 3y) - (8 \times -4)$

$= 3y^2 - 4y - 24y + 32$

Combine the like terms ($-4y$ and $-24y$):

$= 3y^2 - 28y + 32$

The product is $3y^2 - 28y + 32$.

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(iii) (2.5l – 0.5m) and (2.5l + 0.5m)

This multiplication is in the form of the identity $(a-b)(a+b) = a^2 - b^2$.

Here, $a = 2.5l$ and $b = 0.5m$.

Product = $(2.5l)^2 - (0.5m)^2$

$= (2.5 \times 2.5 \times l \times l) - (0.5 \times 0.5 \times m \times m)$

$= 6.25l^2 - 0.25m^2$

The product is $6.25l^2 - 0.25m^2$.

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(iv) (a + 3b) and (x + 5)

Product = $(a + 3b)(x + 5)$

$= a(x + 5) + 3b(x + 5)$

$= (a \times x) + (a \times 5) + (3b \times x) + (3b \times 5)$

$= ax + 5a + 3bx + 15b$

There are no like terms to combine.

The product is $ax + 5a + 3bx + 15b$.

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(v) (2pq + 3q²) and (3pq – 2q²)

Product = $(2pq + 3q^2)(3pq - 2q^2)$

$= 2pq(3pq - 2q^2) + 3q^2(3pq - 2q^2)$

$= (2pq \times 3pq) - (2pq \times 2q^2) + (3q^2 \times 3pq) - (3q^2 \times 2q^2)$

$= 6p^2q^2 - 4pq^3 + 9pq^3 - 6q^4$

Combine the like terms ($-4pq^3$ and $+9pq^3$):

$= 6p^2q^2 + 5pq^3 - 6q^4$

The product is $6p^2q^2 + 5pq^3 - 6q^4$.

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(vi) $\left( \frac{3}{4}a^2 + 3b^2 \right)$ and $4\left( a^2 - \frac{2}{3}b^2 \right)$

First, simplify the second expression by multiplying 4 inside:

$4\left( a^2 - \frac{2}{3}b^2 \right) = 4a^2 - \frac{8}{3}b^2$

Now, multiply the two binomials:

Product = $\left( \frac{3}{4}a^2 + 3b^2 \right) \left( 4a^2 - \frac{8}{3}b^2 \right)$

$= \frac{3}{4}a^2 \left( 4a^2 - \frac{8}{3}b^2 \right) + 3b^2 \left( 4a^2 - \frac{8}{3}b^2 \right)$

$= \left(\frac{3}{4}a^2 \times 4a^2\right) - \left(\frac{3}{4}a^2 \times \frac{8}{3}b^2\right) + \left(3b^2 \times 4a^2\right) - \left(3b^2 \times \frac{8}{3}b^2\right)$

$= 3a^4 - \left(\frac{24}{12}a^2b^2\right) + 12a^2b^2 - \left(\frac{24}{3}b^4\right)$

$= 3a^4 - 2a^2b^2 + 12a^2b^2 - 8b^4$

Combine the like terms ($-2a^2b^2$ and $+12a^2b^2$):

$= 3a^4 + 10a^2b^2 - 8b^4$

The product is $3a^4 + 10a^2b^2 - 8b^4$.

To find the product of the binomials, we use the distributive property. Each term of the first binomial is multiplied by each term of the second binomial, and then the like terms are combined.

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(i) (5 – 2x) (3 + x)

Product = $(5 - 2x)(3 + x)$

$= 5(3 + x) - 2x(3 + x)$

$= (5 \times 3) + (5 \times x) - (2x \times 3) - (2x \times x)$

$= 15 + 5x - 6x - 2x^2$

Combine the like terms ($+5x$ and $-6x$):

$= 15 - x - 2x^2$

Arranging in standard form (decreasing powers of x):

$= -2x^2 - x + 15$

The product is $-2x^2 - x + 15$.

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(ii) (x + 7y) (7x – y)

Product = $(x + 7y)(7x - y)$

$= x(7x - y) + 7y(7x - y)$

$= (x \times 7x) - (x \times y) + (7y \times 7x) - (7y \times y)$

$= 7x^2 - xy + 49xy - 7y^2$

Combine the like terms ($-xy$ and $+49xy$):

$= 7x^2 + 48xy - 7y^2$

The product is $7x^2 + 48xy - 7y^2$.

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(iii) (a² + b) (a + b²)

Product = $(a^2 + b)(a + b^2)$

$= a^2(a + b^2) + b(a + b^2)$

$= (a^2 \times a) + (a^2 \times b^2) + (b \times a) + (b \times b^2)$

$= a^3 + a^2b^2 + ab + b^3$

There are no like terms to combine.

The product is $a^3 + a^2b^2 + ab + b^3$.

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(iv) (p² – q²) (2p + q)

Product = $(p^2 - q^2)(2p + q)$

$= p^2(2p + q) - q^2(2p + q)$

$= (p^2 \times 2p) + (p^2 \times q) - (q^2 \times 2p) - (q^2 \times q)$

$= 2p^3 + p^2q - 2pq^2 - q^3$

There are no like terms to combine.

The product is $2p^3 + p^2q - 2pq^2 - q^3$.

(i) (x² – 5) (x + 5) + 25

First, we multiply the binomials $(x^2 - 5)$ and $(x+5)$ using the distributive property.

$(x^2 - 5)(x + 5) = x^2(x + 5) - 5(x + 5)$

$= (x^2 \times x) + (x^2 \times 5) - (5 \times x) - (5 \times 5)$

$= x^3 + 5x^2 - 5x - 25$

Now, we add the remaining term, 25, to this result.

Expression = $(x^3 + 5x^2 - 5x - 25) + 25$

$= x^3 + 5x^2 - 5x - 25 + 25$

$= x^3 + 5x^2 - 5x$

The simplified expression is $x^3 + 5x^2 - 5x$.

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(ii) (a² + 5) (b³ + 3) + 5

First, multiply the binomials $(a^2 + 5)$ and $(b^3 + 3)$.

$(a^2 + 5)(b^3 + 3) = a^2(b^3 + 3) + 5(b^3 + 3)$

$= (a^2 \times b^3) + (a^2 \times 3) + (5 \times b^3) + (5 \times 3)$

$= a^2b^3 + 3a^2 + 5b^3 + 15$

Now, add the remaining term, 5, to this result.

Expression = $(a^2b^3 + 3a^2 + 5b^3 + 15) + 5$

$= a^2b^3 + 3a^2 + 5b^3 + 20$

The simplified expression is $a^2b^3 + 3a^2 + 5b^3 + 20$.

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(iii) (t + s²) (t² – s)

We multiply the two binomials using the distributive property.

$(t + s^2)(t^2 - s) = t(t^2 - s) + s^2(t^2 - s)$

$= (t \times t^2) - (t \times s) + (s^2 \times t^2) - (s^2 \times s)$

$= t^3 - ts + s^2t^2 - s^3$

The simplified expression is $t^3 + s^2t^2 - ts - s^3$.

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(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)

We simplify each part of the expression first.

Part 1: $(a + b)(c - d) = ac - ad + bc - bd$

Part 2: $(a - b)(c + d) = ac + ad - bc - bd$

Part 3: $2(ac + bd) = 2ac + 2bd$

Now, we add the results of all three parts:

$(ac - ad + bc - bd) + (ac + ad - bc - bd) + (2ac + 2bd)$

Group the like terms:

$(ac + ac + 2ac) + (-ad + ad) + (bc - bc) + (-bd - bd + 2bd)$

$= 4ac + 0 + 0 + 0$

$= 4ac$

The simplified expression is $4ac$.

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(v) (x + y)(2x + y) + (x + 2y)(x – y)

We simplify each product separately.

Part 1: $(x + y)(2x + y) = x(2x+y) + y(2x+y) = 2x^2 + xy + 2xy + y^2 \ $$ = 2x^2 + 3xy + y^2$

Part 2: $(x + 2y)(x - y) = x(x-y) + 2y(x-y) = x^2 - xy + 2xy - 2y^2 \ $$ = x^2 + xy - 2y^2$

Now, we add the two results:

$(2x^2 + 3xy + y^2) + (x^2 + xy - 2y^2)$

Group and combine like terms:

$(2x^2 + x^2) + (3xy + xy) + (y^2 - 2y^2)$

$= 3x^2 + 4xy - y^2$

The simplified expression is $3x^2 + 4xy - y^2$.

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(vi) (x + y)(x² – xy + y²)

We multiply the binomial with the trinomial.

$= x(x^2 - xy + y^2) + y(x^2 - xy + y^2)$

$= (x^3 - x^2y + xy^2) + (x^2y - xy^2 + y^3)$

Group and combine like terms:

$= x^3 + (-x^2y + x^2y) + (xy^2 - xy^2) + y^3$

$= x^3 + 0 + 0 + y^3$

$= x^3 + y^3$

The simplified expression is $x^3 + y^3$.

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(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y

First, multiply the binomial and the trinomial.

$(1.5x - 4y)(1.5x + 4y + 3)$

$= 1.5x(1.5x + 4y + 3) - 4y(1.5x + 4y + 3)$

$= (2.25x^2 + 6xy + 4.5x) - (6xy + 16y^2 + 12y)$

$= 2.25x^2 + 6xy + 4.5x - 6xy - 16y^2 - 12y$

$= 2.25x^2 - 16y^2 + 4.5x - 12y$

Now, include the rest of the original expression:

$(2.25x^2 - 16y^2 + 4.5x - 12y) - 4.5x + 12y$

Group and combine like terms:

$= 2.25x^2 - 16y^2 + (4.5x - 4.5x) + (-12y + 12y)$

$= 2.25x^2 - 16y^2 + 0 + 0$

$= 2.25x^2 - 16y^2$

The simplified expression is $2.25x^2 - 16y^2$.

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(viii) (a + b + c)(a + b – c)

We can group the terms as $[(a+b) + c][(a+b) - c]$.

This is in the form of the identity $(x+y)(x-y) = x^2 - y^2$, where $x = (a+b)$ and $y = c$.

$= (a+b)^2 - c^2$

Now, we expand $(a+b)^2$ using the identity $(a+b)^2 = a^2 + 2ab + b^2$.

$= a^2 + 2ab + b^2 - c^2$

The simplified expression is $a^2 + b^2 - c^2 + 2ab$.

Solution:

The Identity (I) is: $(a + b)^2 = a^2 + 2ab + b^2$

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(i) (2x + 3y)²

We can use Identity (I) by letting $a = 2x$ and $b = 3y$.

$(2x + 3y)^2 = (2x)^2 + 2(2x)(3y) + (3y)^2$

Now, we simplify each term:

$(2x)^2 = 2^2 \times x^2 = 4x^2$

$2(2x)(3y) = (2 \times 2 \times 3)(x \times y) = 12xy$

$(3y)^2 = 3^2 \times y^2 = 9y^2$

Combining these terms, we get:

$(2x + 3y)^2 = 4x^2 + 12xy + 9y^2$

The result is $4x^2 + 12xy + 9y^2$.

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(ii) 103²

We can write 103 as a sum of two numbers, for example, $100 + 3$.

$103^2 = (100 + 3)^2$

Now, we can apply Identity (I) with $a = 100$ and $b = 3$.

$(100 + 3)^2 = (100)^2 + 2(100)(3) + (3)^2$

Simplify each term:

$(100)^2 = 10000$

$2(100)(3) = 600$

$(3)^2 = 9$

Adding these results:

$10000 + 600 + 9 = 10609$

The result is $10609$.

Solution:

The Identity (II) is: $(a - b)^2 = a^2 - 2ab + b^2$

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(i) (4p – 3q)²

We use Identity (II) by setting $a = 4p$ and $b = 3q$.

$(4p - 3q)^2 = (4p)^2 - 2(4p)(3q) + (3q)^2$

Simplify each term:

$(4p)^2 = 4^2 \times p^2 = 16p^2$

$2(4p)(3q) = (2 \times 4 \times 3)(p \times q) = 24pq$

$(3q)^2 = 3^2 \times q^2 = 9q^2$

Combining the terms:

$(4p - 3q)^2 = 16p^2 - 24pq + 9q^2$

The result is $16p^2 - 24pq + 9q^2$.

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(ii) (4.9)²

We can express 4.9 as a difference, for example, $5 - 0.1$.

$(4.9)^2 = (5 - 0.1)^2$

Now, we apply Identity (II) with $a = 5$ and $b = 0.1$.

$(5 - 0.1)^2 = (5)^2 - 2(5)(0.1) + (0.1)^2$

Simplify each term:

$(5)^2 = 25$

$2(5)(0.1) = 10 \times 0.1 = 1$

$(0.1)^2 = 0.01$

Performing the subtraction and addition:

$25 - 1 + 0.01 = 24 + 0.01 = 24.01$

The result is $24.01$.

Solution:

The Identity (III) is the difference of squares: $(a + b)(a - b) = a^2 - b^2$

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(i) $\left( \frac{3}{2}m + \frac{2}{3}n \right)\left( \frac{3}{2}m - \frac{2}{3}n \right)$

This expression is in the form of $(a+b)(a-b)$, where $a = \frac{3}{2}m$ and $b = \frac{2}{3}n$.

Using the identity, the product is $a^2 - b^2$.

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

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

$= \frac{9}{4}m^2 - \frac{4}{9}n^2$

The result is $\frac{9}{4}m^2 - \frac{4}{9}n^2$.

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(ii) 983² – 17²

This expression is in the form of $a^2 - b^2$, where $a = 983$ and $b = 17$.

Using the identity, this is equal to $(a+b)(a-b)$.

$983^2 - 17^2 = (983 + 17)(983 - 17)$

Calculate the terms in the brackets:

$983 + 17 = 1000$

$983 - 17 = 966$

Now, multiply the results:

$(1000)(966) = 966000$

The result is $966,000$.

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(iii) 194 × 206

We can write 194 as $(200 - 6)$ and 206 as $(200 + 6)$.

So, the product is $(200 - 6)(200 + 6)$.

This is in the form of $(a - b)(a + b)$, where $a = 200$ and $b = 6$.

Using the identity, this is equal to $a^2 - b^2$.

$(200 - 6)(200 + 6) = (200)^2 - (6)^2$

Calculate the squares:

$(200)^2 = 40000$

$(6)^2 = 36$

Now, perform the subtraction:

$40000 - 36 = 39964$

The result is $39,964$.

Solution:

The Identity is: $(x + a)(x + b) = x^2 + (a + b)x + ab$

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(i) 501 × 502

We can express the numbers in the form $(x+a)$ and $(x+b)$.

$501 = 500 + 1$

$502 = 500 + 2$

So, the product is $(500 + 1)(500 + 2)$.

Here, $x = 500$, $a = 1$, and $b = 2$.

Using the identity:

$(500 + 1)(500 + 2) = (500)^2 + (1 + 2)(500) + (1)(2)$

Simplify each term:

$(500)^2 = 250000$

$(1 + 2)(500) = 3 \times 500 = 1500$

$(1)(2) = 2$

Add the results:

$250000 + 1500 + 2 = 251502$

The result is $251,502$.

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(ii) 95 × 103

We can express the numbers in the form $(x+a)$ and $(x+b)$. A good choice for $x$ is 100.

$95 = 100 - 5$

$103 = 100 + 3$

So, the product is $(100 - 5)(100 + 3)$.

Here, $x = 100$, $a = -5$, and $b = 3$.

Using the identity:

$(100 - 5)(100 + 3) = (100)^2 + (-5 + 3)(100) + (-5)(3)$

Simplify each term:

$(100)^2 = 10000$

$(-5 + 3)(100) = -2 \times 100 = -200$

$(-5)(3) = -15$

Combine the results:

$10000 - 200 - 15 = 9800 - 15 = 9785$

The result is $9,785$.

(i) (x + 3) (x + 3)

This is of the form $(a+b)(a+b) = (a+b)^2$. We use the identity $(a+b)^2 = a^2 + 2ab + b^2$.

Here, $a=x$ and $b=3$.

$(x+3)^2 = (x)^2 + 2(x)(3) + (3)^2 = x^2 + 6x + 9$.

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(ii) (2y + 5) (2y + 5)

This is of the form $(a+b)^2$. Here, $a=2y$ and $b=5$.

$(2y+5)^2 = (2y)^2 + 2(2y)(5) + (5)^2 = 4y^2 + 20y + 25$.

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(iii) (2a – 7) (2a – 7)

This is of the form $(a-b)(a-b) = (a-b)^2$. We use the identity $(a-b)^2 = a^2 - 2ab + b^2$.

Here, $a=2a$ and $b=7$.

$(2a-7)^2 = (2a)^2 - 2(2a)(7) + (7)^2 = 4a^2 - 28a + 49$.

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(iv) $\left( 3a - \frac{1}{2} \right)\left( 3a - \frac{1}{2} \right)$

This is of the form $(a-b)^2$. Here, $a=3a$ and $b=\frac{1}{2}$.

$\left(3a - \frac{1}{2}\right)^2 = (3a)^2 - 2(3a)\left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)^2 = 9a^2 - 3a + \frac{1}{4}$.

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(v) (1.1m – 0.4) (1.1m + 0.4)

This is of the form $(a-b)(a+b)$. We use the identity $(a-b)(a+b) = a^2 - b^2$.

Here, $a=1.1m$ and $b=0.4$.

$(1.1m)^2 - (0.4)^2 = 1.21m^2 - 0.16$.

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(vi) (a² + b²) (– a² + b²)

We can rearrange the terms to fit the identity: $(b^2 + a^2)(b^2 - a^2)$.

This is of the form $(x+y)(x-y) = x^2 - y^2$. Here, $x=b^2$ and $y=a^2$.

$(b^2)^2 - (a^2)^2 = b^4 - a^4$.

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(vii) (6x – 7) (6x + 7)

This is of the form $(a-b)(a+b) = a^2 - b^2$.

Here, $a=6x$ and $b=7$.

$(6x)^2 - (7)^2 = 36x^2 - 49$.

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(viii) (– a + c) (– a + c)

This can be written as $(c-a)(c-a) = (c-a)^2$.

Using the identity $(x-y)^2 = x^2 - 2xy + y^2$, where $x=c$ and $y=a$.

$(c)^2 - 2(c)(a) + (a)^2 = c^2 - 2ac + a^2$.

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(ix) $\left( \frac{x}{2} + \frac{3y}{4} \right)\left( \frac{x}{2} + \frac{3y}{4} \right)$

This is of the form $(a+b)^2$. Here, $a=\frac{x}{2}$ and $b=\frac{3y}{4}$.

$\left(\frac{x}{2}\right)^2 + 2\left(\frac{x}{2}\right)\left(\frac{3y}{4}\right) + \left(\frac{3y}{4}\right)^2 = \frac{x^2}{4} + \frac{3xy}{4} + \frac{9y^2}{16}$.

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(x) (7a – 9b) (7a – 9b)

This is of the form $(a-b)^2$. Here, the first term is $7a$ and the second term is $9b$.

$(7a)^2 - 2(7a)(9b) + (9b)^2 = 49a^2 - 126ab + 81b^2$.

The identity to be used is: $(X + a)(X + b) = X^2 + (a+b)X + ab$.

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(i) (x + 3) (x + 7)

Here, $X=x$, $a=3$, $b=7$.

Product = $x^2 + (3+7)x + (3)(7)$

$= x^2 + 10x + 21$.

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(ii) (4x + 5) (4x + 1)

Here, $X=4x$, $a=5$, $b=1$.

Product = $(4x)^2 + (5+1)(4x) + (5)(1)$

$= 16x^2 + 6(4x) + 5$

$= 16x^2 + 24x + 5$.

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(iii) (4x – 5) (4x – 1)

Here, $X=4x$, $a=-5$, $b=-1$.

Product = $(4x)^2 + (-5 + (-1))(4x) + (-5)(-1)$

$= 16x^2 + (-6)(4x) + 5$

$= 16x^2 - 24x + 5$.

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(iv) (4x + 5) (4x – 1)

Here, $X=4x$, $a=5$, $b=-1$.

Product = $(4x)^2 + (5 + (-1))(4x) + (5)(-1)$

$= 16x^2 + (4)(4x) - 5$

$= 16x^2 + 16x - 5$.

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(v) (2x + 5y) (2x + 3y)

Here, $X=2x$, $a=5y$, $b=3y$.

Product = $(2x)^2 + (5y+3y)(2x) + (5y)(3y)$

$= 4x^2 + (8y)(2x) + 15y^2$

$= 4x^2 + 16xy + 15y^2$.

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(vi) (2a² + 9) (2a² + 5)

Here, $X=2a^2$, $a=9$, $b=5$.

Product = $(2a^2)^2 + (9+5)(2a^2) + (9)(5)$

$= 4a^4 + 14(2a^2) + 45$

$= 4a^4 + 28a^2 + 45$.

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(vii) (xyz – 4) (xyz – 2)

Here, $X=xyz$, $a=-4$, $b=-2$.

Product = $(xyz)^2 + (-4 + (-2))(xyz) + (-4)(-2)$

$= x^2y^2z^2 + (-6)(xyz) + 8$

$= x^2y^2z^2 - 6xyz + 8$.

(i) (b – 7)²

Using the identity $(a-b)^2 = a^2 - 2ab + b^2$, with $a=b$ and $b=7$.

$(b-7)^2 = (b)^2 - 2(b)(7) + (7)^2 = b^2 - 14b + 49$.

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(ii) (xy + 3z)²

Using the identity $(a+b)^2 = a^2 + 2ab + b^2$, with $a=xy$ and $b=3z$.

$(xy+3z)^2 = (xy)^2 + 2(xy)(3z) + (3z)^2 = x^2y^2 + 6xyz + 9z^2$.

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(iii) (6x² – 5y)²

Using the identity $(a-b)^2 = a^2 - 2ab + b^2$, with $a=6x^2$ and $b=5y$.

$(6x^2-5y)^2 = (6x^2)^2 - 2(6x^2)(5y) + (5y)^2 = 36x^4 - 60x^2y + 25y^2$.

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(iv) $\left( \frac{2}{3} m \;+\;\frac{3}{2}n\right)^2$

Using the identity $(a+b)^2 = a^2 + 2ab + b^2$, with $a=\frac{2}{3}m$ and $b=\frac{3}{2}n$.

$\left(\frac{2}{3}m\right)^2 + 2\left(\frac{2}{3}m\right)\left(\frac{3}{2}n\right) + \left(\frac{3}{2}n\right)^2$

$= \frac{4}{9}m^2 + 2(mn) + \frac{9}{4}n^2 = \frac{4}{9}m^2 + 2mn + \frac{9}{4}n^2$.

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(v) (0.4p – 0.5q)²

Using the identity $(a-b)^2 = a^2 - 2ab + b^2$, with $a=0.4p$ and $b=0.5q$.

$(0.4p)^2 - 2(0.4p)(0.5q) + (0.5q)^2$

$= 0.16p^2 - 2(0.2pq) + 0.25q^2 = 0.16p^2 - 0.4pq + 0.25q^2$.

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(vi) (2xy + 5y)²

Using the identity $(a+b)^2 = a^2 + 2ab + b^2$, with $a=2xy$ and $b=5y$.

$(2xy)^2 + 2(2xy)(5y) + (5y)^2 = 4x^2y^2 + 20xy^2 + 25y^2$.

(i) (a² – b²)²

Using the identity $(x-y)^2 = x^2 - 2xy + y^2$, with $x=a^2$ and $y=b^2$.

$(a^2)^2 - 2(a^2)(b^2) + (b^2)^2 = a^4 - 2a^2b^2 + b^4$.

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(ii) (2x + 5)² – (2x – 5)²

This is of the form $a^2 - b^2$, where $a = (2x+5)$ and $b = (2x-5)$. We use the identity $a^2-b^2 = (a+b)(a-b)$.

$= [(2x+5) + (2x-5)][(2x+5) - (2x-5)]$

$= [2x+5+2x-5][2x+5-2x+5]$

$= [4x][10] = 40x$.

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(iii) (7m – 8n)² + (7m + 8n)²

We expand both terms using $(a-b)^2$ and $(a+b)^2$.

$= (49m^2 - 112mn + 64n^2) + (49m^2 + 112mn + 64n^2)$

Combine like terms:

$= (49m^2+49m^2) + (-112mn+112mn) + (64n^2+64n^2)$

$= 98m^2 + 0 + 128n^2 = 98m^2 + 128n^2$.

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(iv) (4m + 5n)² + (5m + 4n)²

Expand both terms:

$= ((4m)^2 + 2(4m)(5n) + (5n)^2) + ((5m)^2 + 2(5m)(4n) + (4n)^2)$

$= (16m^2 + 40mn + 25n^2) + (25m^2 + 40mn + 16n^2)$

Combine like terms:

$= (16m^2+25m^2) + (40mn+40mn) + (25n^2+16n^2)$

$= 41m^2 + 80mn + 41n^2$.

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(v) (2.5p – 1.5q)² – (1.5p – 2.5q)²

Using the identity $a^2 - b^2 = (a+b)(a-b)$, with $a = (2.5p-1.5q)$ and $b = (1.5p-2.5q)$.

$= [(2.5p-1.5q) + (1.5p-2.5q)][(2.5p-1.5q) - (1.5p-2.5q)]$

$= [2.5p-1.5q+1.5p-2.5q][2.5p-1.5q-1.5p+2.5q]$

$= [4p - 4q][p + q] = 4(p-q)(p+q) = 4(p^2 - q^2)$.

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(vi) (ab + bc)² – 2ab²c

First, expand $(ab+bc)^2$ using $(x+y)^2 = x^2 + 2xy + y^2$.

$= (ab)^2 + 2(ab)(bc) + (bc)^2 - 2ab^2c$

$= a^2b^2 + 2ab^2c + b^2c^2 - 2ab^2c$

Combine like terms:

$= a^2b^2 + b^2c^2$.

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(vii) (m² – n²m)² + 2m³n²

First, expand $(m^2 - n^2m)^2$ using $(x-y)^2 = x^2 - 2xy + y^2$.

$= (m^2)^2 - 2(m^2)(n^2m) + (n^2m)^2 + 2m^3n^2$

$= m^4 - 2m^3n^2 + n^4m^2 + 2m^3n^2$

Combine like terms:

$= m^4 + n^4m^2$.

To show that the given equations are true, we will simplify the Left Hand Side (L.H.S.) and show that it is equal to the Right Hand Side (R.H.S.).

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(i) (3x + 7)² – 84x = (3x – 7)²

L.H.S. = $(3x + 7)^2 - 84x$

Expand $(3x+7)^2$ using $(a+b)^2 = a^2+2ab+b^2$:

L.H.S. = $((3x)^2 + 2(3x)(7) + 7^2) - 84x$

$= (9x^2 + 42x + 49) - 84x$

$= 9x^2 + 42x - 84x + 49$

$= 9x^2 - 42x + 49$

Now, let's look at the R.H.S. = $(3x - 7)^2$.

Expand using $(a-b)^2 = a^2-2ab+b^2$:

R.H.S. = $(3x)^2 - 2(3x)(7) + 7^2$

$= 9x^2 - 42x + 49$

Since L.H.S. = R.H.S., the identity is shown to be true.

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(ii) (9p – 5q)² + 180pq = (9p + 5q)²

L.H.S. = $(9p - 5q)^2 + 180pq$

Expand $(9p-5q)^2$ using $(a-b)^2 = a^2-2ab+b^2$:

L.H.S. = $((9p)^2 - 2(9p)(5q) + (5q)^2) + 180pq$

$= (81p^2 - 90pq + 25q^2) + 180pq$

$= 81p^2 - 90pq + 180pq + 25q^2$

$= 81p^2 + 90pq + 25q^2$

Now, let's expand the R.H.S. = $(9p + 5q)^2$ using $(a+b)^2 = a^2+2ab+b^2$:

R.H.S. = $(9p)^2 + 2(9p)(5q) + (5q)^2$

$= 81p^2 + 90pq + 25q^2$

Since L.H.S. = R.H.S., the identity is shown to be true.

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(iii) $\left( \frac{4}{3} m \;-\;\frac{3}{4}n\right)^2$ + 2mn = $\frac{16}{9}$m² + $\frac{9}{16}$n²

L.H.S. = $\left( \frac{4}{3}m - \frac{3}{4}n \right)^2 + 2mn$

Expand the squared term using $(a-b)^2 = a^2-2ab+b^2$:

L.H.S. = $\left(\left(\frac{4}{3}m\right)^2 - 2\left(\frac{4}{3}m\right)\left(\frac{3}{4}n\right) + \left(\frac{3}{4}n\right)^2\right) + 2mn$

$= \left(\frac{16}{9}m^2 - 2mn + \frac{9}{16}n^2\right) + 2mn$

$= \frac{16}{9}m^2 - 2mn + 2mn + \frac{9}{16}n^2$

$= \frac{16}{9}m^2 + \frac{9}{16}n^2$

This is equal to the R.H.S. Hence, the identity is shown to be true.

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(iv) (4pq + 3q)² – (4pq – 3q)² = 48pq²

The L.H.S. is of the form $a^2 - b^2$, where $a = (4pq+3q)$ and $b = (4pq-3q)$.

Using the identity $a^2-b^2 = (a+b)(a-b)$:

L.H.S. = $[(4pq+3q) + (4pq-3q)][(4pq+3q) - (4pq-3q)]$

$= [4pq+3q+4pq-3q][4pq+3q-4pq+3q]$

$= [8pq][6q]$

$= 48pq^2$

This is equal to the R.H.S. Hence, the identity is shown to be true.

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(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0

We simplify each term on the L.H.S. using the identity $(x-y)(x+y) = x^2-y^2$.

Part 1: $(a - b)(a + b) = a^2 - b^2$

Part 2: $(b - c)(b + c) = b^2 - c^2$

Part 3: $(c - a)(c + a) = c^2 - a^2$

Now, add these three results:

L.H.S. = $(a^2 - b^2) + (b^2 - c^2) + (c^2 - a^2)$

$= a^2 - b^2 + b^2 - c^2 + c^2 - a^2$

Group like terms:

$= (a^2 - a^2) + (-b^2 + b^2) + (-c^2 + c^2)$

$= 0 + 0 + 0 = 0$

This is equal to the R.H.S. Hence, the identity is shown to be true.

(i) 71²

We can write 71 as $(70+1)$.

$71^2 = (70+1)^2$. Using $(a+b)^2 = a^2+2ab+b^2$:

$= 70^2 + 2(70)(1) + 1^2 = 4900 + 140 + 1 = 5041$.

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(ii) 99²

We can write 99 as $(100-1)$.

$99^2 = (100-1)^2$. Using $(a-b)^2 = a^2-2ab+b^2$:

$= 100^2 - 2(100)(1) + 1^2 = 10000 - 200 + 1 = 9801$.

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(iii) 102²

We can write 102 as $(100+2)$.

$102^2 = (100+2)^2$. Using $(a+b)^2 = a^2+2ab+b^2$:

$= 100^2 + 2(100)(2) + 2^2 = 10000 + 400 + 4 = 10404$.

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(iv) 998²

We can write 998 as $(1000-2)$.

$998^2 = (1000-2)^2$. Using $(a-b)^2 = a^2-2ab+b^2$:

$= 1000^2 - 2(1000)(2) + 2^2 = 1000000 - 4000 + 4 = 996004$.

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(v) 5.2²

We can write 5.2 as $(5+0.2)$.

$5.2^2 = (5+0.2)^2$. Using $(a+b)^2 = a^2+2ab+b^2$:

$= 5^2 + 2(5)(0.2) + (0.2)^2 = 25 + 2 + 0.04 = 27.04$.

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(vi) 297 × 303

We can write 297 as $(300-3)$ and 303 as $(300+3)$.

This is of the form $(a-b)(a+b) = a^2-b^2$.

$= 300^2 - 3^2 = 90000 - 9 = 89991$.

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(vii) 78 × 82

We can write 78 as $(80-2)$ and 82 as $(80+2)$.

This is of the form $(a-b)(a+b) = a^2-b^2$.

$= 80^2 - 2^2 = 6400 - 4 = 6396$.

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(viii) 8.9²

We can write 8.9 as $(9-0.1)$.

$8.9^2 = (9-0.1)^2$. Using $(a-b)^2 = a^2-2ab+b^2$:

$= 9^2 - 2(9)(0.1) + (0.1)^2 = 81 - 1.8 + 0.01 = 79.2 + 0.01 = 79.21$.

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(ix) 10.5 × 9.5

We can write this as $(10+0.5) \times (10-0.5)$.

This is of the form $(a+b)(a-b) = a^2-b^2$.

$= 10^2 - (0.5)^2 = 100 - 0.25 = 99.75$.

We use the identity $a^2 - b^2 = (a+b)(a-b)$ to solve the following problems.

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(i) 51² – 49²

Here, $a = 51$ and $b = 49$.

So, $51^2 - 49^2 = (51 + 49)(51 - 49)$

$= (100)(2)$

$= 200$

The value is 200.

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(ii) (1.02)² – (0.98)²

Here, $a = 1.02$ and $b = 0.98$.

So, $(1.02)^2 - (0.98)^2 = (1.02 + 0.98)(1.02 - 0.98)$

$= (2.00)(0.04)$

$= 0.08$

The value is 0.08.

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(iii) 153² – 147²

Here, $a = 153$ and $b = 147$.

So, $153^2 - 147^2 = (153 + 147)(153 - 147)$

$= (300)(6)$

$= 1800$

The value is 1800.

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(iv) 12.1² – 7.9²

Here, $a = 12.1$ and $b = 7.9$.

So, $12.1^2 - 7.9^2 = (12.1 + 7.9)(12.1 - 7.9)$

$= (20.0)(4.2)$

$= 84$

The value is 84.

We use the identity $(x + a)(x + b) = x^2 + (a+b)x + ab$ to solve the following problems.

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(i) 103 × 104

We can write this as $(100 + 3)(100 + 4)$.

Here, $x=100$, $a=3$, and $b=4$.

Product = $(100)^2 + (3+4)(100) + (3)(4)$

$= 10000 + (7)(100) + 12$

$= 10000 + 700 + 12 = 10712$.

The value is 10,712.

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(ii) 5.1 × 5.2

We can write this as $(5 + 0.1)(5 + 0.2)$.

Here, $x=5$, $a=0.1$, and $b=0.2$.

Product = $(5)^2 + (0.1 + 0.2)(5) + (0.1)(0.2)$

$= 25 + (0.3)(5) + 0.02$

$= 25 + 1.5 + 0.02 = 26.52$.

The value is 26.52.

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(iii) 103 × 98

We can write this as $(100 + 3)(100 - 2)$.

Here, $x=100$, $a=3$, and $b=-2$.

Product = $(100)^2 + (3 + (-2))(100) + (3)(-2)$

$= 10000 + (1)(100) - 6$

$= 10000 + 100 - 6 = 10094$.

The value is 10,094.

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(iv) 9.7 × 9.8

We can write this as $(10 - 0.3)(10 - 0.2)$.

Here, $x=10$, $a=-0.3$, and $b=-0.2$.

Product = $(10)^2 + (-0.3 + (-0.2))(10) + (-0.3)(-0.2)$

$= 100 + (-0.5)(10) + 0.06$

$= 100 - 5 + 0.06 = 95.06$.

The value is 95.06.