# Complete NCERT solutions for class 8 Maths Chapter 9 Algebraic Expressions

Solution for Exercise 9.1

1.Identify the terms, their coefficients for each of the following expressions :
(i) $5xy{z}^{2}–3zy$

(ii) $1+x+{x}^{2}$

(iii) $4{x}^{2}{y}^{2}–4{x}^{2}{y}^{2}{z}^{2}+{z}^{2}$

(iv) $3–pq+qr?rp$

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

(vi)$0.3a–0.6ab+0.5b$

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## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

(i) We have the expression $5xy{z}^{2}–3zy$, the terms are $5xy{z}^{2}$ and $?3zy$.
Coefficient of $xy{z}^{2}$ in the term $5xy{z}^{2}$ is 5.

Coefficient of $zy$ in the term $–3yz$ is $–3$.

(ii)We have the expression $1+x+{x}^{2}$, the terms are 1, x and ${x}^{2}$.

Coefficient of the term 1 is 1.

Coefficient of x in the term x is 1.

Coefficient of ${x}^{2}$ in the term ${x}^{2}$ is 1.

(iii) We have the expression $4xy–4xyz+z$ , the terms are $4{x}^{2}{y}^{2},–4{x}^{2}{y}^{2}{z}^{2}$ and ${z}^{2}$.

Coefficient of ${x}^{2}{y}^{2}$ in the term $4{x}^{2}{y}^{2}$ is 4.

Coefficient of ${x}^{2}{y}^{2}{z}^{2}$ in the term $–4{x}^{2}{y}^{2}{z}^{2}$ is – 4.

Coefficient of ${z}^{2}$ in the term ${z}^{2}$ is 1.

(iv) We have the expression $3–pq+qr–rp$, the terms are 3, – pq, qr and – rp.

Coefficient of the term 3 is 3.

Coefficient of pq in the term – pq is -1.

Coefficient of qr in the term qr is 1.

Coefficient of rp in the term – rp is -1.

(v) We have the expression $\frac{x}{2}+\frac{y}{2}?xy$, the terms are $\frac{x}{2},\frac{7}{2}$ and $–xy$.

Coefficient of x in the term $\frac{x}{2}$ is $\frac{1}{2}$

Coefficient of y in the term $\frac{y}{2}$ is $\frac{1}{2}$

Coefficient of xy in the term $–xy$ is -1.

(vi) In the expression $0.3a?0.6ab+0.5b$, the terms are $0.3a,–006ab$and $0.5b$.

Coefficient of a in the term 0.3a is 0.3.

Coefficient of ab in the term – 0.6ab is – 0.6.

Coefficient of b in the term 0.5b is 0.5.

2.Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories? $x+y,1000,x+{x}^{2}+{x}^{3}+{x}^{4},7+y+5x,2y–3{y}^{2},2y–3{y}^{2}+4{y}^{3}$,$5x–4y+3xy,4z–15{z}^{2},ab+bc+cd+da,pqr,{p}^{2}q+p{q}^{2},2p+2q.$

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## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

The calssification of the given expressions are:

Monomials : 1000, pqr

Binomials : $x+y,2y–3{y}^{2},4z?15{z}^{2},{p}^{2}q+p{q}^{2},2p+2q.$

Trinomials : $7+y+5x,2y–3{y}^{2}+4{y}^{3},5x–4y+3x.$

And the polynomials which don't have any category:

$x+{x}^{2}+{x}^{3}+{x}^{4},ab+be+cd+da.$

(i) $ab–be,be–ca,ea–ab$

(ii)$a–b+ab,b?c+be,c–a+ac$

(iii) $2{p}^{2}{q}^{2}–3pq+4,5+7pq–3{p}^{2}{q}^{2}$

(iv)${l}^{2}+{m}^{2},{m}^{2}+{n}^{2},{n}^{2}+{l}^{2},2lm+2mn+2nl$

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## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

(i) We have the expressions as:$ab–bc,bc–ca,ca–ab$

$\left(ab–bc\right)+\left(bc–ca\right)+\left(ca–ab\right)$

$?ab–bc+bc–ca+ca–ab$

$?\left(ab–ab\right)+\left(bc–bc\right)+\left(ca–ca\right)\phantom{\rule{1em}{0ex}}$ [Placing the like terms together]

$?0+0+0$

$?0$

(ii) We have the expressions as:$a–b+ab,b–c+bc,c–a+ac$

$\left(a–b+ab\right)+\left(b–c+bc\right)+\left(c–a+ac\right)$

$?a–b+ab+b–c+bc+c–a+ac$

$?\left(a–a\right)+\left(b–b\right)+\left(c–c\right)+ab+bc+ac\phantom{\rule{1em}{0ex}}$[Placing all the like terms together]

$?0+0+0+ab+bc+ac$

$?ab+bc+ac$

(iii)We have the expressions as:$2{p}^{2}{q}^{2}–3pq+4,5+7pq–3{p}^{2}{q}^{2}$

Adding these we get: $2{p}^{2}{q}^{2}–3pq+4+5+7pq–3{p}^{2}{q}^{2}$

$?\left(2{p}^{2}{q}^{2}–3{p}^{2}{q}^{2}\right)+\left(–3pq+7pq\right)+\left(4+5\right)\phantom{\rule{1em}{0ex}}$[Placing like terms together]

$??{p}^{2}{q}^{2}+4pq+9$

(iv)We have the expressions as:${l}^{2}+{m}^{2},{m}^{2}+{n}^{2},{n}^{2}+{l}^{2},2lm+2mn+2nl$
Adding these we get: $\left({l}^{2}+{m}^{2}\right)+\left({m}^{2}+{n}^{2}\right)+\left({n}^{2}+{l}^{2}\right)+\left(2lm+2mn+2nl\right)$

$?\left({l}^{2}+{l}^{2}\right)+\left({m}^{2}+{m}^{2}\right)+\left({n}^{2}+{n}^{2}\right)+\left(2lm+2mn+2nl\right)\phantom{\rule{1em}{0ex}}$[Placing like terms together]

$?\left(2{l}^{2}\right)+\left(2{m}^{2}\right)+\left(2{n}^{2}\right)+\left(2lm+2mn+2nl$

$?2\left({l}^{2}+{m}^{2}+{n}^{2}+lm+mn+nl\right)$

4.(a) Subtract $4a–7ab+3b+12$ from $12a–9ab+5b–3$
(b) Subtract $3xy+5yz–7zx$ from $5xy–2yz–2zx+10xyz$

(c) Subtract $4{p}^{2}q–3pq+5p{q}^{2}–8p+7q–10$ from $18–3p–11q+5pq–2p{q}^{2}+5{p}^{2}q$

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## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

(a)We have the expressions to subtract:$4a–7ab+3b+12$ from $12a–9ab+5b–3$
Subtracting these we get: $\left(12a–9ab+5b–3\right)?\left(4a–7ab+3b+12\right)$

$\left(12a–9ab+5b–3\right)+\left(?4a+7ab?3b?12\right)\phantom{\rule{1em}{0ex}}$[Taking (-) inside the second bracket]

$?\left(12a?4a\right)+\left(–9ab+7ab\right)+\left(+5b?3b\right)+\left(–3?12\right)\phantom{\rule{1em}{0ex}}$[Placing like terms together]

$?\left(8a\right)+\left(?2ab\right)+\left(2b\right)+\left(?15\right)$

$?\left(8a?2ab+2b?15\right)$

(b)We have the expressions to subtract:$3xy+5yz–7zx$ from $5xy–2yz–2zx+10xyz$
Subtracting these we get: $\left(5xy–2yz–2zx+10xyz\right)?\left(3xy+5yz–7zx\right)$

$\left(5xy–2yz–2zx+10xyz\right)+\left(?3xy?5yz+7zx\right)\phantom{\rule{1em}{0ex}}$[Taking (-) inside the second bracket]

$?\left(5xy?3xy\right)+\left(–2yz?5yz\right)+\left(–2zx+7zx\right)+\left(+10xyz\right)\phantom{\rule{1em}{0ex}}$[Placing like terms together]

$?\left(2xy\right)+\left(?7yz\right)+\left(5zx\right)+\left(10xyz\right)$

$?\left(2xy?7yz+5zx+10xyz\right)$

(c)We have the expressions to subtract:$4{p}^{2}q–3pq+5p{q}^{2}–8p+7q–10$ from $18–3p–11q+5pq–2p{q}^{2}+5{p}^{2}q$
Subtracting these we get: $\left(18–3p–11q+5pq–2p{q}^{2}+5{p}^{2}q\right)?\left(4{p}^{2}q–3pq+5p{q}^{2}–8p+7q–10\right)$

$?\left(18–3p–11q+5pq–2p{q}^{2}+5{p}^{2}q\right)+\left(?4{p}^{2}q+3pq?5p{q}^{2}+8p?7q+10\right)\phantom{\rule{1em}{0ex}}$[Taking (-) inside the second bracket]

$?\left(+18+10\right)+\left(?3p+8p\right)+\left(?11q?7q\right)+\left(5pq+3pq\right)+\left(?2p{q}^{2}?5p{q}^{2}\right)+\left(+5{p}^{2}q?4{p}^{2}q\right)\phantom{\rule{1em}{0ex}}$[Placing like terms together]

$?\left(+28\right)+\left(+5p\right)+\left(?18q\right)+\left(8pq\right)+\left(?7p{q}^{2}\right)+\left(+{p}^{2}q\right)$

$?\left(+28+5p?18q+8pq?7p{q}^{2}+{p}^{2}q\right)$

4.Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(i) $5a,3{a}^{2},7{a}^{4}$

(ii) $2p,4q,8r$

(iii) $xy,2{x}^{2}y,2x{y}^{2}$

(iv) $a,2b,3c$

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## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

(i)Given that, length = 5a, breadth = $3{a}^{2}$, height = $7{a}^{4}$

?Volume of the box =$l×b×h=5a×3{a}^{2}×7{a}^{4}=105{a}^{7}\phantom{\rule{thickmathspace}{0ex}}$cu. units

(ii)Given that, length = 2p, breadth = 4q, height = 8r

?Volume of the box =$l×b×h=2p×4q×8r=64pqr\phantom{\rule{thickmathspace}{0ex}}$ cu. units

(iii)Given that, length = xy, breadth =$2{x}^{2}y$, height = 2xy^2\)

Volume of the box =$l×b×h=xy×2{x}^{2}y×2x{y}^{2}=\left(1×2×2\right)×xy×{x}^{2}y×x{y}^{2}=4{x}^{4}{y}^{4}\phantom{\rule{thickmathspace}{0ex}}$cu. units

(iv)Given that, length = a, breadth = 2b, height = 3c

?Volume of the box = $l×b×h=a×2b×3c=\left(1×2×3\right)abc=6abc\phantom{\rule{thickmathspace}{0ex}}$ cu. units

Solution for Exercise 9.2

1.Find the product of the following pairs of monomials :
(i) 4, 7p

(ii) -4p, 7p

(iii) – 4p, 7pq

(iv) 4p3 , – 3p

(v) 4p, 0

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## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

(i)Multiplying the given expressions: $4x×7p=\left(4×7\right)×p=28p$

(ii) Multiplying the given expressions:$–4p×7p=\left(?4×7\right)×\left(p×p\right)$

$=?28{p}^{\left(1+1\right)}=–28{p}^{2}$

(iii)Multiplying the given expressions: $?4p×7pq=\left(?4×7\right)×\left(p×p×q\right)$

$=?28{p}^{\left(1+1\right)}q=–28{p}^{2}q$

(iv)Multiplying the given expressions: $4p3×–3p=\left(4×–3\right)×\left({p}^{3}×p\right)$

$=–12{p}^{\left(3+1\right)}==?12{p}^{4}$

(v) Multiplying the given expressions:$4p×0=\left(4×0\right)×p=0×p=0$

2.Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.
$\left(p,q\right);\left(10m,5n\right);\left(20{x}^{2},5{y}^{2}\right);\left(4x,3{x}^{2}\right);\left(3mn,4np\right)$

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## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

(i) Given that: Length = p units and breadth = q units

Area of the rectangle = $length×breadth=p×q=pq$ sq units

(ii)Given that: Length = 10 m units, breadth = 5n units

?Area of the rectangle =$length×breadth=10m×5n=\left(10×5\right)×m×n=50mn\phantom{\rule{thickmathspace}{0ex}}$sq units

(iii) Given that: Length = 20x2 units, breadth = 5y2 units

?Area of the rectangle = length × breadth = 20x2 × 5y2 = (20 × 5) × x2 × y2 = 100x2y2 sq units

(iv)Given that: Length = 4x units, breadth = 3x2 units

?Area of the rectangle = $length×breadth=4x×3{x}^{2}=\left(4×3\right)×x×{x}^{2}=12{x}^{3}\phantom{\rule{thickmathspace}{0ex}}$ sq units

(v) Given that: Length = 3mn units, breadth = 4np units

?Area of the rectangle = $length×breadth=3mn×4np=\left(3×4\right)×mn×np=12m{n}^{2}p\phantom{\rule{thickmathspace}{0ex}}$ sq units

3.Complete the table of products : Open in new tab

## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

Completed table is as under : 5. Obtain the product of

(i) xy, yz, zx

(ii) a, -a2, a3

(iii) 2, 4y, 8y2, 16y3

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

(v) m, -mn, mnp

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## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

(i) Product=$xy×yz×zx={x}^{2}{y}^{2}{z}^{2}$

(ii) Product=$a×\left(?{a}^{2}\right)×{a}^{3}=\left(?a{\right)}^{6}$

(iii)Product=$2×4y×8{y}^{2}×16{y}^{3}=\left(2×4×8×16\right)×y×{y}^{2}×{y}^{3}=1024{y}^{6}$

(iv)Product=$a×2b×3c×6abc=\left(1×2×3×6\right)×a×b×c×abc=36{a}^{2}{b}^{2}{c}^{2}$

(v)Product=$m×\left(?mn\right)×mnp=\left[1×\left(?1\right)×1\right]m×mn×mnp=?{m}^{3}{n}^{2}p$

Solution for Exercise 9.3

1.Carry out the multiplication of the expressions in each of the following pairs:
(i) $4p,q+r$

(ii) $ab,a–b$

(iii) $a+b,7{a}^{2}{b}^{2}$

(iv) ${a}^{2}–9,4a$

(v) $pq+qr+rp,0$

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## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

Multiplication of the expressions are as follows:
(i) $4p×\left(q+r\right)=\left(4p×q\right)+\left(4p×r\right)=4pq+4pr$

(ii) $ab,a–b=ab×\left(a–b\right)=\left(ab×a\right)–\left(ab×b\right)={a}^{2}b–a{b}^{2}$

(iii) $\left(a+b\right)×7{a}^{2}{b}^{2}=\left(a×7{a}^{2}{b}^{2}\right)+\left(b×7{a}^{2}{b}^{2}\right)=7{a}^{3}{b}^{2}+7{a}^{2}{b}^{3}$

(iv) $\left(a2–9\right)×4a=\left({a}^{2}×4a\right)–\left(9×4a\right)=4{a}^{3}–36a$

(v) $\left(pq+qr+rp\right)×0=0$

2. Complete the table. Open in new tab

## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

Products of the given expressions in the table are done as follows:
.(i) $a×\left(b+c+d\right)=\left(a×b\right)+\left(a×c\right)+\left(a×d\right)=ab+ac+ad$

(ii) $\left(x+y–5\right)\left(5xy\right)=\left(x×5xy\right)+\left(y×5xy\right)–\left(5×5xy\right)=5{x}^{2}y+5x{y}^{2}–25xy$

(iii) $p×\left(6{p}^{2}–7p+5\right)=\left(p×6{p}^{2}\right)–\left(p×7p\right)+\left(p×5\right)=6{p}^{3}–7{p}^{2}+5p$

(iv) $4{p}^{2}{q}^{2}×\left({p}^{2}–{q}^{2}\right)=4{p}^{2}{q}^{2}×{p}^{2}–4{p}^{2}{q}^{2}×{q}^{2}=4{p}^{4}{q}^{2}–4{p}^{2}{q}^{4}$

(v) $\left(a+b+c\right)×\left(abc\right)=\left(a×abc\right)+\left(b×abc\right)+\left(c×abc\right)={a}^{2}bc+a{b}^{2}c+ab{c}^{2}$

Therefore, the completed table is as follows: 3.Find the products
(i)$\left({a}^{2}\right)×\left(2{a}^{22}\right)×\left(4{a}^{26}\right)$

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

(iii)$\left(?\frac{10}{3}p{q}^{3}\right)×\left(\frac{6}{5}{p}^{3}q\right)$

(iv)$x×{x}^{2}×{x}^{3}×{x}^{4}$

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## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

(i)$\left({a}^{2}\right)×\left(2{a}^{22}\right)×\left(4{a}^{26}\right)$

$=\left(1×2×4\right)×{a}^{2+22+26}=8{a}^{50}$

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

$=\left(\frac{2}{3}×\left(?\frac{9}{10}\right)\right)×{x}^{\left(1+2}\right){y}^{\left(1+2\right)}$

(iii)$\left(?\frac{10}{3}p{q}^{3}\right)×\left(\frac{6}{5}{p}^{3}q\right)$

$=\left(?\frac{10}{3}×\frac{6}{5}\right)×{p}^{\left(1+3\right)}{q}^{\left(1+3\right)}$

$=?4{p}^{4}{q}^{4}$

(iv)$x×{x}^{2}×{x}^{3}×{x}^{4}$

$$= x^{(1+2+3+4)}=x^{10$$

4.(a) Simplify: $3x\left(4x–5\right)+3$ and find its values for (i) x = 3 (ii) x = $\frac{1}{2}$.
(b) Simplify: $a\left({a}^{2}+a+1\right)+5$ and find its value for (i) a = 0 (ii) a = 1 (iii) a = -1

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## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

(a) Given the expression: $3x\left(4x–5\right)+3=4x×3x–5×3x+3=12{x}^{2}–15x+3$
(i) So for x = 3, we have

$12×\left(3\right)2–15×3+3=12×9–45+3=108–42=66$

(ii)For x=$\frac{1}{2}$, we have:

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

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

$?3?\frac{15}{2}+3$

$?\frac{6?15+6}{2}=\frac{12?15}{2}=?\frac{3}{2}$

(b) We have $a\left({a}^{2}+a+1\right)+5$

$=\left({a}^{2}×a\right)+\left(a×a\right)+\left(1×a\right)+5$

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

(i) For a = 0, we have:

$=\left(0{\right)}^{3}+\left(0{\right)}^{2}+\left(0\right)+5=5$

(ii) For a = 1, we have:

$=\left(1{\right)}^{3}+\left(1{\right)}^{2}+\left(1\right)+5=1+1+1+5=8$

(iii) For a = -1, we have:

$=\left(?1{\right)}^{3}+\left(?1{\right)}^{2}+\left(?1\right)+5=?1+1–1+5=4$

5.(a) $Add:p\left(p–q\right),q\left(q–r\right)andr\left(r–p\right)$
(b) $Add:2x\left(z–x–y\right)and2y\left(z–y–x\right)$

(c) $Subtract:3l\left(l–4m+5n\right)from4l\left(10n–3m+2l\right)$

(d) $Subtract:3a\left(a+b+c\right)–2b\left(a–b+c\right)from4c\left(?a+b+c\right)$

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## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

(a) $p\left(p–q\right)+q\left(q–r\right)+r\left(r–p\right)$

=$\left(p×p\right)–\left(p×q\right)+\left(q×q\right)–\left(q×r\right)+\left(r×r\right)–\left(r×p\right)$[Opening the brackets]

= ${p}^{2}–pq+{q}^{2}–qr+{r}^{2}–rp$

=${p}^{2}+{q}^{2}+{r}^{2}–pq–qr–rp$

(b) $2x\left(z–x–y\right)+2y\left(z–y–x\right)$

= $\left(2x×z\right)–\left(2x×x\right)–\left(2x×y\right)+\left(2y×z\right)–\left(2y×y\right)–\left(2y×x\right)$[Opening the brackets]

= $2xz–2{x}^{2}–2xy+2yz–2{y}^{2}–2xy$

$=?2{x}^{2}–2{y}^{2}+2xz+2yz–4xy$

$=?2{x}^{2}–2{y}^{2}–4xy+2yz+2xz$

(c) $4l\left(10n–3m+2l\right)–3l\left(l–4m+5n\right)$

$=\left(4l×10n\right)–\left(4l×3m\right)+\left(4l×2l\right)–\left(3l×l\right)–\left(3l×?4m\right)–\left(3l××5n\right)$[Opening the brackets]

$=40ln–12lm+8{l}^{2}–3{l}^{2}+12lm–15ln$

$=\left(40ln–15ln\right)+\left(?12lm+12lm\right)+\left(8{l}^{2}–3{l}^{2}\right)$

$=25ln+0+5{l}^{2}$

$=25ln+5{l}^{2}$

$=5{l}^{2}+25ln$

(d)$\left[4c\left(?a+b+c\right)\right]–\left[3a\left(a+b+c\right)–2b\left(a–b+c\right)\right]$

$=\left(?4ac+4bc+4{c}^{2}\right)–\left(3{a}^{2}+3ab+3ac–2ab+2{b}^{2}–2bc\right)$[Opening the brackets]

$=?4ac+4bc+4{c}^{2}–3{a}^{2}–3ab–3ac+2ab–2{b}^{2}+2bc$

$=?3{a}^{2}–2{b}^{2}+4{c}^{2}–ab+6bc–7ac$

Solution for Exercise 9.4

1.Multiply the binomials:
(i) $\left(2x+5\right)$and $\left(4x–3\right)$

(ii) $\left(y–8\right)$ and $\left(3y–4\right)$

(iii) $\left(2.5l–0.5m\right)$ and $\left(2.5l+0.5m\right)$

(iv) $\left(a+3b\right)$ and $\left(x+5\right)$

(v) $\left(2pq+3{q}^{2}\right)$ and $\left(3pq–2{q}^{2}\right)$

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

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## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

(i)Multiplying the given expressions as:$\left(2x+5\right)×\left(4x–3\right)$

$=2x×\left(4x–3\right)+5×\left(4x–3\right)$

$=\left(2x×4x\right)–\left(3×2x\right)+\left(5×4x\right)–\left(5×3\right)$

$=8{x}^{2}–6x+20x–15$

$=8{x}^{2}+14x–15$

(ii)Multiplying the given expressions as: $\left(y–8\right)×\left(3y–4\right)$

$=y×\left(3y–4\right)–8×\left(3y–4\right)$

$=\left(y×3y\right)–\left(y×4\right)–\left(8×3y\right)+\left(?8×?4\right)$

$=3{y}^{2}–4y–24y+32$

$=3{y}^{2}–28y+32$

(iii)Multiplying the given expressions as: $\left(2.5l–0.5m\right)×\left(2.5l+0.5m\right)$

$=\left(2.5l×2.5l\right)+\left(2.5l×0.5m\right)–\left(0.5m×2.5l\right)–\left(0.5m×0.5m\right)$

$=6.25{l}^{2}+1.25ml–1.25ml–0.25{m}^{2}$

$=6.25{l}^{2}+0–0.25{m}^{2}$

$=6.25{l}^{2}–0.25{m}^{2}$

(iv)Multiplying the given expressions as: $\left(a+3b\right)×\left(x+5\right)$

$=a×\left(x+5\right)+36×\left(x+5\right)$

$=\left(a×x\right)+\left(a×5\right)+\left(36×x\right)+\left(36×5\right)$

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

(v) Multiplying the given expressions as:$\left(2pq+3{q}^{2}\right)×\left(3pq–2{q}^{2}\right)$

$=2pq×\left(3pq–2{q}^{2}\right)+3{q}^{2}\left(3pq–2{q}^{2}\right)$

$=\left(2pq×3pq\right)–\left(2pq×2{q}^{2}\right)+\left(3{q}^{2}×3pq\right)–\left(3{q}^{2}×2{q}^{2}\right)$

$=6{p}^{2}{q}^{2}–4p{q}^{3}+9p{q}^{3}–6{q}^{4}$

$=6{p}^{2}{q}^{2}+5p{q}^{3}–6{q}^{4}$

(vi)Multiplying the given expressions as:$\left(\frac{3}{4}{a}^{2}+3{b}^{2}\right)×4\left({a}^{2}–\frac{2}{3}{b}^{2}\right)$

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

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

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

$=3{a}^{4}?2{a}^{2}{b}^{2}+12{a}^{2}{b}^{2}?8{b}^{4}$

$=3{a}^{4}+10{a}^{2}{b}^{2}?8{b}^{4}$

2.Find the product:
(i) $\left(5–2x\right)\left(3+x\right)$

(ii)$\left(x+7y\right)\left(7x–y\right)$

(iii) $\left({a}^{2}+b\right)\left(a+{b}^{2}\right)$

(iv)$\left({p}^{2}–{q}^{2}\right)\left(2p+q\right)$

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## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

(i)Product of the expression:$\left(5–2x\right)\left(3+x\right)$

$=5\left(3+x\right)–2x\left(3+x\right)$

$=\left(5×3\right)+\left(5×x\right)–\left(2x×3\right)–\left(2x×x\right)$

$=15+5x–6x–2{x}^{2}$

(ii)Product of the expression: $\left(x+7y\right)\left(7x–y\right)$

$=x\left(7x–y\right)+7y\left(7x–y\right)$

$=\left(x×7x\right)–\left(x×y\right)+\left(7y×7x\right)–\left(7y×y\right)$

$=7{x}^{2}–xy+49xy–7{y}^{2}$

$=7{x}^{2}+48xy–7{y}^{2}$

(iii)Product of the expression: $\left({a}^{2}+b\right)\left(a+{b}^{2}\right)$

$={a}^{2}\left(a+{b}^{2}\right)+b\left(a+{b}^{2}\right)$

$=\left({a}^{2}×a\right)+\left({a}^{2}×{b}^{2}\right)+\left(b×a\right)+\left(b×{b}^{2}\right)$

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

(iv)Product of the expression:$\left({p}^{2}–{q}^{2}\right)\left(2p+q\right)$

$={p}^{2}\left(2p+q\right)–{q}^{2}\left(2p+q\right)$

$=\left({p}^{2}×2p\right)+\left({p}^{2}×q\right)–\left({q}^{2}×2p\right)–\left({q}^{2}×q\right)$

$=2{p}^{3}+{p}^{2}q–2p{q}^{2}–{q}^{3}$

3.Simplify: (i) $\left({x}^{2}–5\right)\left(x+5\right)+25$

(ii) $\left({a}^{2}+5\right)\left({b}^{3}+3\right)+5$

(iii) $\left(t+{s}^{2}\right)\left({t}^{2}–s\right)$

(iv)$\left(a+b\right)\left(c–d\right)+\left(a–b\right)\left(c+d\right)+2\left(ac+bd\right)$

(v)$\left(x+y\right)\left(2x+y\right)+\left(x+2y\right)\left(x–y\right)$

(vi) $\left(x+y\right)\left({x}^{2}–xy+{y}^{2}\right)$

(vii)$\left(1.5x–4y\right)\left(1.5x+4y+3\right)–4.5x+12y$

(viii)$\left(a+b+c\right)\left(a+b–c\right)$

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## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

(i)Simplifying the expression, we have: $\left({x}^{2}–5\right)\left(x+5\right)+25$

$={x}^{2}×\left(x+5\right)+5×\left(x+5\right)+25$

$={x}^{3}+5{x}^{2}–5x–25+25$

$={x}^{3}+5{x}^{2}–5x+0$

$={x}^{3}+5{x}^{2}–5x$

(ii)Simplifying the expression, we have: $\left({a}^{2}+5\right)\left({b}^{3}+3\right)+5$

$={a}^{2}×\left({b}^{3}+3\right)+5×\left({b}^{3}+3\right)+5$

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

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

(iii)Simplifying the expression, we have:$\left(t+{s}^{2}\right)\left({t}^{2}–s\right)$

$=t×\left({t}^{2}–s\right)+{s}^{2}×\left({t}^{2}–s\right)$

$={t}^{3}–st+{s}^{2}{t}^{2}–{s}^{3}$

$={t}^{3}+{s}^{2}{t}^{2}–st–{s}^{3}$

(iv) Simplifying the expression, we have:$\left(a+b\right)\left(c–d\right)+\left(a–b\right)\left(c+d\right)+2\left(ac+bd\right)$

$=a×\left(c–d\right)+b×\left(c–d\right)+a×\left(c+d\right)–b×\left(c+d\right)+2ac+2bd$

$=ac–ad+bc–bd+ac+ad–bc–bd+2ac+2bd$

$=ac+ac+2ac+bc–bc–ad+ad–bd–bd+2bd$

$=4ac+0+0+0$

$=4ac$

(v) Simplifying the expression, we have:$\left(x+y\right)\left(2x+y\right)+\left(x+2y\right)\left(x–y\right)$

$=x×\left(2x+y\right)+y×\left(2x+y\right)+x×\left(x–y\right)+2y×\left(x–y\right)$

$=2{x}^{2}+xy+2xy+{y}^{2}+{x}^{2}–xy+2xy–2{y}^{2}$

$=2{x}^{2}+{x}^{2}+xy+2xy–xy+2xy+{y}^{2}–2{y}^{2}$

$=3{x}^{2}+4xy–{y}^{2}$

(vi)Simplifying the expression, we have:$\left(x+y\right)\left({x}^{2}–xy+{y}^{2}\right)$

$=x×\left({x}^{2}–xy+{y}^{2}\right)+y×\left({x}^{2}–xy+{y}^{2}\right)$

$={x}^{3}–{x}^{2}y+{x}^{2}y+x{y}^{2}–x{y}^{2}+{y}^{3}$

$={x}^{3}–0+0+{y}^{3}$

$={x}^{3}+{y}^{3}$

(vii)Simplifying the expression, we have:$\left(1.5x–4y\right)\left(1.5x+4y+3\right)–4.5x.+12y$

$=1.5x×\left(1.5x+4y+3\right)–4y×\left(1.5x+4y+3\right)–4.5x+12y$

$=2.25{x}^{2}+6xy+4.5x–6xy–16{y}^{2}–12y–4.5x+12y$

$=2.25{x}^{2}+6xy–6xy+4.5x–4.5x+12y–12y–16{y}^{2}$

$=2.25{x}^{2}+0+0+0–16y62$

$=2.25{x}^{2}–16{y}^{2}$

(viii)Simplifying the expression, we have: $\left(a+b+c\right)\left(a+b–c\right)$

$=a\left(a+b–c\right)+b\left(a+b–c\right)+c\left(a+b–c\right)$

$={a}^{2}+ab–ac+ab+{b}^{2}–bc+ac+bc–{c}^{2}$

$={a}^{2}+ab+ab–bc+bc–ac+ac+{b}^{2}–{c}^{2}$

$={a}^{2}+2ab+{b}^{2}–{c}^{2}+0+0$

$={a}^{2}+2ab+{b}^{2}–{c}^{2}$

Solution for Exercise 9.5

1.Use a suitable identity to get each of the following products:
(i) $\left(x+3\right)\left(x+3\right)$

(ii) $\left(2y+5\right)\left(2y+5\right)$

(iii) $\left(2a–7\right)\left(2a–7\right)$

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

(v)$\left(1.1m–0.4\right)\left(1.1m+0.4\right)$

(vi)$\left(a2+b2\right)\left(?a2+b2\right)$

(vii) $\left(6x–7\right)\left(6x+7\right)$

(viii) $\left(?a+c\right)\left(?a+c\right)$

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

(x) $\left(7a–9b\right)\left(7a–9b\right)$

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## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

(i)Given: $\left(x+3\right)\left(x+3\right)$
$=\left(x+3{\right)}^{2}$

$=\left(x{\right)}^{2}+2×x×3+\left(3{\right)}^{2}\phantom{\rule{1em}{0ex}}$[?$\left(a+b{\right)}^{2}={a}^{2}+2ab+{b}^{2}$]

$={x}^{2}+6x+9$

(ii)Given: $\left(2y+5\right)\left(2y+5\right)$
$=\left(2y+5{\right)}^{2}$

$=\left(2y{\right)}^{2}+2×2y×5+\left(5{\right)}^{2}\phantom{\rule{1em}{0ex}}$[?$\left(a+b{\right)}^{2}={a}^{2}+2ab+{b}^{2}$]

$=4{y}^{2}+20y+25$

(iii)Given: $\left(2a?7\right)\left(2a?7\right)$
$=\left(2a?7{\right)}^{2}$

$=\left(2a{\right)}^{2}?2×\left(2a\right)×7+\left(7{\right)}^{2}\phantom{\rule{1em}{0ex}}$[?$\left(a?b{\right)}^{2}={a}^{2}?2ab+{b}^{2}$]

$=4{a}^{2}?28a+49$

(iv)Given: $\left(3a?\frac{1}{2}\right)\left(3a?\frac{1}{2}\right)$
$=\left(3a?\frac{1}{2}{\right)}^{2}$

$=\left(3a{\right)}^{2}?2×\left(3a\right)×\left(\frac{1}{2}\right)+\left(\frac{1}{2}{\right)}^{2}\phantom{\rule{1em}{0ex}}$[?$\left(a?b{\right)}^{2}={a}^{2}?2ab+{b}^{2}$]

$=9{a}^{2}?3a+\frac{1}{4}$

(v)Given: $\left(1.1m?0.4\right)\left(1.1m+0.4\right)$
$=\left(\left(1.1m{\right)}^{2}?\left(0.4{\right)}^{2}\phantom{\rule{1em}{0ex}}$[?$\left(a?b\right)\left(a+b\right)={a}^{2}?{b}^{2}$]

$=1.21{a}^{2}?0.16$

(vi)Given: $\left({a}^{2}+{b}^{2}\right)\left(?{a}^{2}+{b}^{2}\right)$
$=\left({b}^{2}+{a}^{2}\right)\left({b}^{2}?{a}^{2}\right)$

$=\left({b}^{2}{\right)}^{2}?\left({a}^{2}{\right)}^{2}\phantom{\rule{1em}{0ex}}$[?$\left(a?b\right)\left(a+b\right)={a}^{2}?{b}^{2}$]

$=\left({b}^{4}?{a}^{4}\right)$

(vii)Given: $\left(6x?7\right)\left(6x+7\right)$
$=\left(6x{\right)}^{2}?\left(7{\right)}^{2}\phantom{\rule{1em}{0ex}}$[?$\left(a?b\right)\left(a+b\right)={a}^{2}?{b}^{2}$]

$=36{x}^{2}?49$

(viii)Given: $\left(?a+c\right)\left(?a+c\right)$
$=\left[\left(?a\right)+c{\right]}^{2}$

$=\left(?a{\right)}^{2}+2×\left(?a\right)×c+\left(c{\right)}^{2}\phantom{\rule{1em}{0ex}}$[?$\left(a+b{\right)}^{2}={a}^{2}+2ab+{b}^{2}$]

$={a}^{2}?2ac+{c}^{2}$

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

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

$=\left(\frac{x}{2}{\right)}^{2}+2×\left(\frac{x}{2}\right)×\left(\frac{3y}{4}\right)+\left(\frac{3y}{4}{\right)}^{2}\phantom{\rule{1em}{0ex}}$[?$\left(a+b{\right)}^{2}={a}^{2}+2ab+{b}^{2}$]

$=\frac{{x}^{2}}{4}+\frac{3}{4}xy+\frac{9{y}^{2}}{16}$

(x)Given: $\left(7a?9b\right)\left(7a?9b\right)$
$=\left(7a?9b{\right)}^{2}$

$=\left(7a{\right)}^{2}?2×\left(7a\right)×\left(9b\right)+\left(9b{\right)}^{2}\phantom{\rule{1em}{0ex}}$[?$\left(a?b{\right)}^{2}={a}^{2}?2ab+{b}^{2}$]

$=49{a}^{2}?126ab+81{b}^{2}$

2.Use the identity $\left(x+a\right)\left(x+b\right)={x}^{2}+\left(a+b\right)x+ab$ to find the following products.
(i) $\left(x+3\right)\left(x+7\right)$

(ii) $\left(4x+5\right)\left(4x+1\right)$

(iii)$\left(4x–5\right)\left(4x–1\right)$

(iv) $\left(4x+5\right)\left(4x–1\right)$

(v) $\left(2x+5y\right)\left(2x+3y\right)$

(vi) $\left(2a2+9\right)\left(2a2+5\right)$

(vii) $\left(xyz–4\right)\left(xyz–2\right)$

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## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

(i)We have: $\left(x+3\right)\left(x+7\right)$

$={x}^{2}+\left(3+7\right)x+3×7$

$={x}^{2}+10x+21$

(ii)We have: $\left(4x+5\right)\left(4x+1\right)$

$=\left(4x{\right)}^{2}+\left(5+1\right)\left(4x\right)+5×1$

$=16{x}^{2}+24x+5$

(iii)We have: $\left(4x?5\right)\left(4x?1\right)$

$=\left(4x{\right)}^{2}?\left(5+1\right)\left(4x\right)+\left(?5\right)×\left(?1\right)$

$=16{x}^{2}?24x+5$

(iv)We have: $\left(4x+5\right)\left(4x?1\right)$

$=\left(4x{\right)}^{2}+\left(5?1\right)\left(4x\right)+\left(5\right)×\left(?1\right)$

$=16{x}^{2}+16x?5$

(v)We have: $\left(2x+5y\right)\left(2x+3y\right)$

$=\left(2x{\right)}^{2}+\left(5y+3y\right)\left(2x\right)+\left(5y\right)×\left(3y\right)$

$=\left(2x{\right)}^{2}+\left(8y\right)\left(2x\right)+\left(5y\right)×\left(3y\right)$

$=4{x}^{2}+16xy+15{y}^{2}$

(vi)We have: $\left(2{a}^{2}+9\right)\left(2{a}^{2}+5\right)$

$=\left(2{a}^{2}{\right)}^{2}+\left(9+5\right)\left(2{a}^{2}\right)+\left(5\right)×\left(9\right)$

$=\left(2{a}^{2}{\right)}^{2}+\left(14\right)\left(2{a}^{2}\right)+45$

$=4{a}^{4}+28{a}^{2}+45$

(vii)We have: $\left(xyz?4\right)\left(xyz?2\right)$

$=\left(xyz{\right)}^{2}?\left(4+2\right)\left(xyz\right)+\left(?4\right)×\left(?2\right)$

$=\left(xyz{\right)}^{2}?\left(6\right)\left(xyz\right)+8$

$={x}^{2}{y}^{2}{z}^{2}?6xyz+8$

3.Find the following squares by using the identities.
(i) $\left(b–7{\right)}^{2}$

(ii) $\left(xy+3z{\right)}^{2}$

(iii) $\left(6{x}^{2}–5y{\right)}^{2}$

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

(v) $\left(0.4p–0.5q{\right)}^{2}$

(vi) $\left(2xy+5y{\right)}^{2}$

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## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

(i)Given: $\left(b–7{\right)}^{2}$

$=\left(b{\right)}^{2}?2\left(b\right)\left(7\right)+\left(7{\right)}^{2}\phantom{\rule{1em}{0ex}}$[using $\left(a?b{\right)}^{2}={a}^{2}?2ab+{b}^{2}$]

$=\left(b{\right)}^{2}?14b+49$

(ii)Given: $\left(xy+3z{\right)}^{2}$

$=\left(xy{\right)}^{2}+2\left(xy\right)\left(3z\right)+\left(3z{\right)}^{2}\phantom{\rule{1em}{0ex}}$[using $\left(a+b{\right)}^{2}={a}^{2}+2ab+{b}^{2}$]

$={x}^{2}{y}^{2}+6xyz+9{y}^{2}$

(iii)Given:$\left(6{x}^{2}–5y{\right)}^{2}$

$=\left(6{x}^{2}{\right)}^{2}?2\left(6{x}^{2}\right)\left(5y\right)+\left(5y{\right)}^{2}\phantom{\rule{1em}{0ex}}$[using $\left(a?b{\right)}^{2}={a}^{2}?2ab+{b}^{2}$]

$=36{x}^{4}?60{x}^{2}y+25{y}^{2}$

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

$=\left(\frac{2}{3}m{\right)}^{2}+2\left(\frac{2}{3}m\right)\left(\frac{3}{2}n\right)+\left(\frac{3}{2}n{\right)}^{2}\phantom{\rule{1em}{0ex}}$[using $\left(a+b{\right)}^{2}={a}^{2}+2ab+{b}^{2}$]

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

(v)Given: $\left(0.4p–0.5q{\right)}^{2}$

$=\left(0.4p{\right)}^{2}?2\left(0.4p\right)\left(0.5q\right)+\left(0.5q{\right)}^{2}\phantom{\rule{1em}{0ex}}$[using $\left(a?b{\right)}^{2}={a}^{2}?2ab+{b}^{2}$]

$=0.16{p}^{2}?0.4pq+0.25{q}^{2}$

(vi)Given: $\left(2xy+5y{\right)}^{2}$

$=\left(2xy{\right)}^{2}+2\left(2xy\right)\left(5y\right)+\left(5y{\right)}^{2}\phantom{\rule{1em}{0ex}}$[using $\left(a+b{\right)}^{2}={a}^{2}+2ab+{b}^{2}$]

$=4{x}^{2}{y}^{2}+20x{y}^{2}+25{y}^{2}$

4.Simplify:
(i)$\left({a}^{2}–{b}^{2}{\right)}^{2}$

(ii) $\left(2x+5{\right)}^{2}–\left(2x–5{\right)}^{2}$

(iii) $\left(7m–8n{\right)}^{2}+\left(7m+8n{\right)}^{2}$

(iv) $\left(4m+5n{\right)}^{2}+\left(5m+4n{\right)}^{2}$

(v) $\left(2.5p–1.5q{\right)}^{2}–\left(1.5p–2.5q{\right)}^{2}$

(vi) $\left(ab+bc{\right)}^{2}–2a{b}^{2}c$

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

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## Best NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions

(i)Given:$\left({a}^{2}?{b}^{2}{\right)}^{2}$
$=\left({a}^{2}{\right)}^{2}?2{a}^{2}{b}^{2}+\left({b}^{2}{\right)}^{2}\phantom{\rule{1em}{0ex}}$[use $\left(a?b{\right)}^{2}={a}^{2}?2ab+{b}^{2}$]

$={a}^{4}?2{a}^{2}{b}^{2}+{b}^{4}$

(ii)Given:$\left(2x+5{\right)}^{2}–\left(2x–5{\right)}^{2}$

$=\left[\left(2x{\right)}^{2}+2\left(2x\right)\left(5\right)+\left(5{\right)}^{2}\right]?\left[\left(2x{\right)}^{2}?2\left(2x\right)\left(5\right)+\left(5{\right)}^{2}\right]$

$=\left[4{x}^{2}+20x+25\right]?\left[4{x}^{2}?20x+25\right]$

$=4{x}^{2}+20x+25?4{x}^{2}+20x?25\right]$

$=\left(4{x}^{2}?4{x}^{2}\right)+\left(+20x+20x\right)+\left(+25?25\right)\phantom{\rule{1em}{0ex}}$[Placing same terms together]

$=40x$

(iii)Given:$\left(7m–8n{\right)}^{2}+\left(7m+8n{\right)}^{2}$

$=\left[\left(7m{\right)}^{2}?2\left(7m\right)\left(8n\right)+\left(8n{\right)}^{2}\right]+\left[\left(7m{\right)}^{2}+2\left(7m\right)\left(8n\right)+\left(8n{\right)}^{2}\right]$

$=\left[49{m}^{2}?112mn+64{n}^{2}\right]+\left[49{m}^{2}+112mn+64{n}^{2}\right]$

$=\left(49{m}^{2}+49{m}^{2}\right)+\left(?112mn+112mn\right)+\left(64{n}^{2}+64{n}^{2}\right)\phantom{\rule{1em}{0ex}}$[Placing same terms together]

$=98{m}^{2}+128{n}^{2}$

(iv)Given:$\left(4m+5n{\right)}^{2}+\left(5m+4n{\right)}^{2}$

$=\left[\left(4m{\right)}^{2}+2\left(4m\right)\left(5n\right)+\left(5n{\right)}^{2}\right]+\left[\left(5m{\right)}^{2}+2\left(5m\right)\left(4n\right)+\left(4n{\right)}^{2}\right]$

$=\left[16{m}^{2}+40mn+25{n}^{2}\right]+\left[25{m}^{2}+40mn+16{n}^{2}\right]$

$=\left(16{m}^{2}+25{m}^{2}\right)+\left(40mn+40mn\right)+\left(25{n}^{2}+16{n}^{2}\right)\phantom{\rule{1em}{0ex}}$[Placing same terms together]

$=41{m}^{2}+80mn+41{n}^{2}$

(v)Given: $\left(2.5p–1.5q{\right)}^{2}–\left(1.5p–2.5q{\right)}^{2}$

$=\left[\left(2.5p{\right)}^{2}?2\left(2.5p\right)\left(1.5q\right)+\left(1.5q{\right)}^{2}\right]?\left[\left(1.5p{\right)}^{2}?2\left(1.5p\right)\left(2.5q\right)+\left(2.5q{\right)}^{2}\right]$

$=\left[6.25{p}^{2}?7.5pq+2.25{q}^{2}\right]?\left[2.25{p}^{2}?7.5pq+6.25{q}^{2}\right]$

$=6.25{p}^{2}?7.5pq+2.25{q}^{2}?2.25{p}^{2}+7.5pq?6.25{q}^{2}$

$=\left(6.25{p}^{2}?2.25{p}^{2}\right)+\left(?7.5pq+7.5pq\right)+\left(2.25{q}^{2}?6.25{q}^{2}\right)\phantom{\rule{1em}{0ex}}$[Placing same terms together]

$=4{p}^{2}?4{q}^{2}$

(vi)Given:$\left(ab+bc{\right)}^{2}–2a{b}^{2}c$

$=\left(ab{\right)}^{2}+2\left(ab\right)\left(bc\right)+\left(bc{\right)}^{2}?2a{b}^{2}c$

$={a}^{2}{b}^{2}+2a{b}^{2}c+{b}^{2}{c}^{2}?2a{b}^{2}c$

$={a}^{2}{b}^{2}+{b}^{2}{c}^{2}$

(vii)Given:$\left({m}^{2}–{n}^{2}m{\right)}^{2}+2{m}^{3}{n}^{2}$
$=\left({m}^{2}{\right)}^{2}?2\left({m}^{2}\right)\left({n}^{2}m\right)+\left({n}^{2}m{\right)}^{2}+2{m}^{3}{n}^{2}$

$={m}^{4}?2{m}^{3}{n}^{2}+{n}^{4}{m}^{2}+2{m}^{3}{n}^{2}$

$={m}^{4}+{n}^{4}{m}^{2}$

5. Show that:
(i)$\left(3x+7{\right)}^{2}–84x=\left(3x–7{\right)}^{2}$

(ii) $\left(9p–5q{\right)}^{2}+180pq=\left(9p+5q{\right)}^{2}$

(iii)