Rs Aggarwal 2019 2020 Solutions for Class 8 Maths Chapter 6 Operations On Algebraic Expressions are provided here with simple step-by-step explanations. These solutions for Operations On Algebraic Expressions are extremely popular among Class 8 students for Maths Operations On Algebraic Expressions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 2020 Book of Class 8 Maths Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rs Aggarwal 2019 2020 Solutions. All Rs Aggarwal 2019 2020 Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

#### Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

________

$5ab$
â€‹

#### Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

_____

#### Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

___________

#### Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

#### Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

#### Page No 84:

On arranging the terms of the given expressions in the descending powers of $x$ and adding column-wise:

#### Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise:

#### Page No 84:

On arranging the terms of the given expressions in the descending powers of $x$ and adding column-wise:

#### Page No 84:

On arranging the terms of the given expressions in the descending powers of $x$ and subtracting:

#### Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

#### Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

#### Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

#### Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

#### Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

#### Page No 84:

On arranging the terms of the given expressions in the descending powers of $x$ and subtracting column-wise:

#### Page No 84:

Arranging the terms of the given expressions in the descending powers of $x$ and subtracting column-wise:

#### Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

#### Page No 84:

Let the required number be $x$.
$\left(3{a}^{2}-6ab-3{b}^{2}-1\right)-x=4{a}^{2}-7ab-4{b}^{2}+1$
$\left(3{a}^{2}-6ab-3{b}^{2}-1\right)-\left(4{a}^{2}-7ab-4{b}^{2}+1\right)=x\phantom{\rule{0ex}{0ex}}$

∴ Required number = $-{a}^{2}+ab+{b}^{2}-2$

#### Page No 84:

Sides of the rectangle are $l$ and $b$.
$l=5{x}^{2}-3{y}^{2}\phantom{\rule{0ex}{0ex}}b={x}^{2}+2xy$
Perimeter of the rectangle is $\left(2l+2b\right)$.

#### Page No 84:

Let be the three sides of the triangle.

∴ Perimeter of the triangle =$\left(a+b+c\right)$

Given perimeter of the triangle = $6{p}^{2}-4p+9$
One side ($a$)  = ${p}^{2}-2p+1$
Other side ($b$) = $3{p}^{2}-5p+3$
Perimeter = $\left(a+b+c\right)$

Thus, the third side is $2{p}^{2}+3p+5$.

#### Page No 87:

By horizontal method:
$\left(5x+7\right)×\left(3x+4\right)\phantom{\rule{0ex}{0ex}}=5x\left(3x+4\right)+7\left(3x+4\right)\phantom{\rule{0ex}{0ex}}=15{x}^{2}+20x+21x+28\phantom{\rule{0ex}{0ex}}=15{x}^{2}+41x+28$

#### Page No 87:

By horizontal method:

$\left(4x+9\right)×\left(x-6\right)\phantom{\rule{0ex}{0ex}}=4x\left(x-6\right)+9\left(x-6\right)\phantom{\rule{0ex}{0ex}}=4{x}^{2}-24x+9x-54\phantom{\rule{0ex}{0ex}}=4{x}^{2}-15x-54$

#### Page No 87:

By horizontal method:

$\left(2x+5\right)×\left(4x-3\right)\phantom{\rule{0ex}{0ex}}=2x\left(4x-3\right)+5\left(4x-3\right)\phantom{\rule{0ex}{0ex}}=8{x}^{2}-6x+20x-15\phantom{\rule{0ex}{0ex}}=8{x}^{2}+14x-15$

#### Page No 87:

By horizontal method:

$\left(3y-8\right)×\left(5y-1\right)\phantom{\rule{0ex}{0ex}}=3y\left(5y-1\right)-8\left(5y-1\right)\phantom{\rule{0ex}{0ex}}=15{y}^{2}-3y-40y+8\phantom{\rule{0ex}{0ex}}=15{y}^{2}-43y+8$

#### Page No 87:

By horizontal method:

$\left(7x+2y\right)×\left(x+4y\right)\phantom{\rule{0ex}{0ex}}=7x\left(x+4y\right)+2y\left(x+4y\right)\phantom{\rule{0ex}{0ex}}=7{x}^{2}+28xy+2xy+8{y}^{2}\phantom{\rule{0ex}{0ex}}=7{x}^{2}+30xy+8{y}^{2}$

#### Page No 87:

By horizontal method:

$\left(9x+5y\right)×\left(4x+3y\right)\phantom{\rule{0ex}{0ex}}9x\left(4x+3y\right)+5y\left(4x+3y\right)\phantom{\rule{0ex}{0ex}}=36{x}^{2}+27xy+20xy+15{y}^{2}\phantom{\rule{0ex}{0ex}}=36{x}^{2}+47xy+15{y}^{2}$

#### Page No 87:

By horizontal method:

$\left(3m-4n\right)×\left(2m-3n\right)\phantom{\rule{0ex}{0ex}}=3m\left(2m-3n\right)-4n\left(2m-3n\right)\phantom{\rule{0ex}{0ex}}=6{m}^{2}-9mn-8mn+12{n}^{2}\phantom{\rule{0ex}{0ex}}=6{m}^{2}-17mn+12{n}^{2}$

#### Page No 87:

By horizontal method:

$\left({x}^{2}-{a}^{2}\right)×\left(x-a\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left(x-a\right)-{a}^{2}\left(x-a\right)\phantom{\rule{0ex}{0ex}}={x}^{3}-a{x}^{2}-{a}^{2}x+{a}^{3}$
i.e $\left({x}^{3}+{a}^{3}\right)-ax\left(x-a\right)$

#### Page No 87:

By horizontal method:

$\left({x}^{2}-{y}^{2}\right)×\left(x+2y\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left(x+2y\right)-{y}^{2}\left(x+2y\right)\phantom{\rule{0ex}{0ex}}={x}^{3}+2{x}^{2}y-x{y}^{2}-2{y}^{3}\phantom{\rule{0ex}{0ex}}i.e\left({x}^{3}-2{y}^{3}\right)+xy\left(2x-y\right)$

#### Page No 87:

By horizontal method:

$\left(3{p}^{2}+{q}^{2}\right)×\left(2{p}^{2}-3{q}^{2}\right)\phantom{\rule{0ex}{0ex}}=3{p}^{2}\left(2{p}^{2}-3{q}^{2}\right)+{q}^{2}\left(2{p}^{2}-3{q}^{2}\right)\phantom{\rule{0ex}{0ex}}=6{p}^{4}-9{p}^{2}{q}^{2}+2{p}^{2}{q}^{2}-3{q}^{4}\phantom{\rule{0ex}{0ex}}i.e6{p}^{4}-7{p}^{2}{q}^{2}-3{q}^{4}$

#### Page No 87:

By horizontal method:

$\left(2{x}^{2}-5{y}^{2}\right)×\left({x}^{2}+3{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=2{x}^{2}\left({x}^{2}+3{y}^{2}\right)-5{y}^{2}\left({x}^{2}+3{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=2{x}^{4}+6{x}^{2}{y}^{2}-5{x}^{2}{y}^{2}-15{y}^{4}\phantom{\rule{0ex}{0ex}}=2{x}^{4}+{x}^{2}{y}^{2}-15{y}^{4}\phantom{\rule{0ex}{0ex}}$

#### Page No 87:

By horizontal method:

$\left({x}^{3}-{y}^{3}\right)×\left({x}^{2}+{y}^{2}\right)\phantom{\rule{0ex}{0ex}}={x}^{3}\left({x}^{2}+{y}^{2}\right)-{y}^{3}\left({x}^{2}+{y}^{2}\right)\phantom{\rule{0ex}{0ex}}={x}^{5}+{x}^{3}{y}^{2}-{x}^{2}{y}^{3}-{y}^{5}\phantom{\rule{0ex}{0ex}}=\left({x}^{5}-{y}^{5}\right)+{x}^{2}{y}^{2}\left(x-y\right)$

#### Page No 87:

By horizontal method:
$\left({x}^{4}+{y}^{4}\right)×\left({x}^{2}-{y}^{2}\right)\phantom{\rule{0ex}{0ex}}={x}^{4}\left({x}^{2}-{y}^{2}\right)+{y}^{4}\left({x}^{2}-{y}^{2}\right)\phantom{\rule{0ex}{0ex}}={x}^{6}-{x}^{4}{y}^{2}+{y}^{4}{x}^{2}-{y}^{6}\phantom{\rule{0ex}{0ex}}=\left({x}^{6}-{y}^{6}\right)-{x}^{2}{y}^{2}\left({x}^{2}-{y}^{2}\right)$

#### Page No 87:

By horizontal method:

#### Page No 87:

By horizontal method:

$\left({x}^{2}-3x+7\right)×\left(2x+3\right)\phantom{\rule{0ex}{0ex}}=2x\left({x}^{2}-3x+7\right)+3\left({x}^{2}-3x+7\right)\phantom{\rule{0ex}{0ex}}=2{x}^{3}-6{x}^{2}+14x+3{x}^{2}-9x+21\phantom{\rule{0ex}{0ex}}=2{x}^{3}-3{x}^{2}+5x+21\phantom{\rule{0ex}{0ex}}$

#### Page No 87:

By horizontal method:
$\left(3{x}^{2}+5x-9\right)×\left(3x-5\right)\phantom{\rule{0ex}{0ex}}=3x\left(3{x}^{2}+5x-9\right)-5\left(3{x}^{2}+5x-9\right)\phantom{\rule{0ex}{0ex}}=9{x}^{3}+15{x}^{2}-27x-15{x}^{2}-25x+45\phantom{\rule{0ex}{0ex}}=9{x}^{3}-52x+45$

#### Page No 87:

By horizontal method:
$\left({x}^{2}-xy+{y}^{2}\right)×\left(x+y\right)\phantom{\rule{0ex}{0ex}}=x\left({x}^{2}-xy+{y}^{2}\right)+y\left({x}^{2}-xy+{y}^{2}\right)\phantom{\rule{0ex}{0ex}}={x}^{3}-{x}^{2}y+{y}^{2}x+{x}^{2}y-x{y}^{2}+{y}^{3}\phantom{\rule{0ex}{0ex}}={x}^{3}+{y}^{3}$

#### Page No 87:

By horizontal method:

$\left({x}^{2}+xy+{y}^{2}\right)×\left(x-y\right)\phantom{\rule{0ex}{0ex}}x\left({x}^{2}+xy+{y}^{2}\right)-y\left({x}^{2}+xy+{y}^{2}\right)\phantom{\rule{0ex}{0ex}}={x}^{3}+{x}^{2}y+x{y}^{2}-{x}^{2}y-x{y}^{2}-{y}^{3}\phantom{\rule{0ex}{0ex}}={x}^{3}-{y}^{3}$

#### Page No 87:

By horizontal method:

$\left({x}^{3}-2{x}^{2}+5\right)×\left(4x-1\right)\phantom{\rule{0ex}{0ex}}=4x\left({x}^{3}-2{x}^{2}+5\right)-1\left({x}^{3}-2{x}^{2}+5\right)\phantom{\rule{0ex}{0ex}}=4{x}^{4}-8{x}^{3}+20x-{x}^{3}+2{x}^{2}-5\phantom{\rule{0ex}{0ex}}=4{x}^{4}-9{x}^{3}+2{x}^{2}+20x-5$

#### Page No 87:

By horizontal method:

$\left(9{x}^{2}-x+15\right)×\left({x}^{2}-3\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left(9{x}^{2}-x+15\right)-3\left(9{x}^{2}-x+15\right)\phantom{\rule{0ex}{0ex}}=9{x}^{4}-{x}^{3}+15{x}^{2}-27{x}^{2}+3x-45\phantom{\rule{0ex}{0ex}}=9{x}^{4}-{x}^{3}-12{x}^{2}+3x-45$

#### Page No 87:

By horizontal method:

$\left({x}^{2}-5x+8\right)×\left({x}^{2}+2\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left({x}^{2}-5x+8\right)+2\left({x}^{2}-5x+8\right)\phantom{\rule{0ex}{0ex}}={x}^{4}-5{x}^{3}+8{x}^{2}+2{x}^{2}-10x+16\phantom{\rule{0ex}{0ex}}={x}^{4}-5{x}^{3}+10{x}^{2}-10x+16$

#### Page No 87:

By horizontal method:

$\left({x}^{3}-5{x}^{2}+3x+1\right)×\left({x}^{2}-3\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left({x}^{3}-5{x}^{2}+3x+1\right)-3\left({x}^{3}-5{x}^{2}+3x+1\right)\phantom{\rule{0ex}{0ex}}={x}^{5}-5{x}^{4}+3{x}^{3}+{x}^{2}-3{x}^{3}+15{x}^{2}-9x-3\phantom{\rule{0ex}{0ex}}={x}^{5}-5{x}^{4}+16{x}^{2}-9x-3$

#### Page No 87:

By horizontal method:

$\left(3x+2y-4\right)×\left(x-y+2\right)\phantom{\rule{0ex}{0ex}}x\left(3x+2y-4\right)-y\left(3x+2y-4\right)+2\left(3x+2y-4\right)\phantom{\rule{0ex}{0ex}}=3{x}^{2}+2xy-4x-3xy-2{y}^{2}+4y+6x+4y-8\phantom{\rule{0ex}{0ex}}=3{x}^{2}-2{y}^{2}-xy+2x+8y-8$

#### Page No 87:

By horizontal method:

$\left({x}^{2}-5x+8\right)×\left({x}^{2}+2x-3\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left({x}^{2}-5x+8\right)+2x\left({x}^{2}-5x+8\right)-3\left({x}^{2}-5x+8\right)\phantom{\rule{0ex}{0ex}}={x}^{4}-5{x}^{3}+8{x}^{2}+2{x}^{3}-10{x}^{2}+16x-3{x}^{2}+15x-24\phantom{\rule{0ex}{0ex}}={x}^{4}-3{x}^{3}-5{x}^{2}+31x-24$

#### Page No 87:

By horizontal method:

$\left(2{x}^{2}+3x-7\right)×\left(3{x}^{2}-5x+4\right)\phantom{\rule{0ex}{0ex}}=2{x}^{2}\left(3{x}^{2}-5x+4\right)+3x\left(3{x}^{2}-5x+4\right)-7\left(3{x}^{2}-5x+4\right)\phantom{\rule{0ex}{0ex}}=6{x}^{4}-10{x}^{3}+8{x}^{2}+9{x}^{3}-15{x}^{2}+12x-21{x}^{2}+35x-28\phantom{\rule{0ex}{0ex}}=6{x}^{4}-{x}^{3}-28{x}^{2}+47x-28$

#### Page No 87:

By horizontal method:

$\left(9{x}^{2}-x+15\right)×\left({x}^{2}-x-1\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left(9{x}^{2}-x+15\right)-x\left(9{x}^{2}-x+15\right)-1\left(9{x}^{2}-x+15\right)\phantom{\rule{0ex}{0ex}}=9{x}^{4}-{x}^{3}+15{x}^{2}-9{x}^{3}+{x}^{2}-15x-9{x}^{2}+x-15\phantom{\rule{0ex}{0ex}}=9{x}^{4}-10{x}^{3}+7{x}^{2}-14x-15$

#### Page No 90:

(i) 24x2y3 by 3xy

Therefore, the quotient is 8xy2.

(ii) 36xyz2 by −9xz

Therefore, the quotient is 4yz.

(iii)

Therefore, the quotient is 6xy.

(iv) −56mnp2 by 7mnp

Therefore, the quotient is −8p.

#### Page No 90:

(i) 5m3 − 30m2 + 45m by 5m

Therefore, the quotient is m2 6m + 9.

(ii) 8x2y2 − 6xy2 + 10x2y3 by 2xy

Therefore, the quotient is 4xy 3y + 5xy2.

(iii) 9x2y − 6xy + 12xy2 by − 3xy

Therefore, the quotient is −3x + 2 4y.

(iv) 12x4 + 8x3 − 6x2 by − 2x2

2-4x+32

Therefore the quotient is −6x2 4x + 3.

#### Page No 90:

Therefore, the quotient is $\left(x-2\right)$ and the remainder is 0.

#### Page No 90:

Therefore, the quotient is $x$−2 and the remainder is 0.

#### Page No 90:

(x2 + 12x + 35) by (x + 7)

Therefore, the quotient is $\left(x+5\right)$ and the remainder is 0.

#### Page No 90:

Therefore, the quotient is $\left(5x-3\right)$ and the remainder is 0.

#### Page No 90:

Therefore, the quotient is and the remainder is 0.

#### Page No 90:

Therefore, the quotient is and the remainder is 7.

#### Page No 90:

Therefore, the quotient is $\left({x}^{2}-x-1\right)$ and the remainder is 1.

#### Page No 90:

Therefore, the quotient is ${x}^{2}$-x+1 and the remainder is 0.

#### Page No 90:

Therefore, the quotient is ( x2 - 3x + 4) and remainder is 0.

#### Page No 90:

Therefore, the quotient is (x-1) and the remainder is 0.

#### Page No 90:

Therefore, the quotient is ( 5x+ 3) and the remainder is (x + 1).

#### Page No 90:

Therefore, the quotient is (x-1) and the remainder is 0.

#### Page No 90:

Therefore, the quotient is ( 4x2+ 3x -2) and the remainder is ( x-1).

(i) We have:

(ii) We have:

(iii) We have:

(iv) We have:

(v) We have:

(vi) We have:

(i) We have:

(ii) We have:

(iii) We have:

(iv) We have:

(v) We have:

(vi) We have:

#### Page No 93:

We shall use the identities (a+b)2 =a2 +b2 +2ab and (a-b)2 =a2 +b2 -2ab.

(i) We have:
$\left(8a+3b{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(8a\right)}^{2}+2×8a×3b+{\left(3b\right)}^{2}\phantom{\rule{0ex}{0ex}}=64{a}^{2}+48ab+9{b}^{2}$

(ii)We have:
$\left(7x+2y{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(7x\right)}^{2}+2×7x×2y+{\left(2y\right)}^{2}\phantom{\rule{0ex}{0ex}}=49{x}^{2}+28xy+4{y}^{2}$

(iii) We have :
$\left(5x+11{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(5x\right)}^{2}+2×5x×11+{\left(11\right)}^{2}\phantom{\rule{0ex}{0ex}}=25{x}^{2}+110x+121$

(iv) We have:
${\left(\frac{a}{2}+\frac{2}{a}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{a}{2}\right)}^{2}+2×\frac{a}{2}×\frac{2}{a}+{\left(\frac{2}{a}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\frac{a}{4}}^{2}+2+\frac{4}{{a}^{2}}$

(v) We have:
${\left(\frac{3x}{4}+\frac{2y}{9}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{3x}{4}\right)}^{2}+2×\frac{3x}{4}×\frac{2y}{9}+{\left(\frac{2y}{9}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\frac{9x}{16}}^{2}+\frac{1}{3}xy+\frac{4{y}^{2}}{81}$

(vi) We have:
$\left(9x-10{\right)}^{2}\phantom{\rule{0ex}{0ex}}{\left(9x\right)}^{2}-2×9x×10+{\left(10\right)}^{2}\phantom{\rule{0ex}{0ex}}=81{x}^{2}-180x+100$

(vii) We have:
$\left({x}^{2}y-y{z}^{2}{\right)}^{2}\phantom{\rule{0ex}{0ex}}{\left({x}^{2}y\right)}^{2}-2×{x}^{2}y×y{z}^{2}+{\left(y{z}^{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}={x}^{4}{y}^{2}-2{x}^{2}{y}^{2}{z}^{2}+{y}^{2}{z}^{4}$

(viii) We have:
${\left(\frac{x}{y}-\frac{y}{x}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{x}{y}\right)}^{2}-2×\frac{x}{y}×\frac{y}{x}+{\left(\frac{y}{x}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{2}}{{y}^{2}}-2+\frac{{y}^{2}}{{x}^{2}}$

(ix) We have:
${\left(3m-\frac{4}{5}n\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(3m\right)}^{2}-2×3m×\frac{4}{5}n+{\left(\frac{4}{5}n\right)}^{2}\phantom{\rule{0ex}{0ex}}=9{m}^{2}-\frac{24mn}{5}+\frac{16}{25}{n}^{2}$

(i) We have:

(ii) We have:

(iii) We have:

(iv) We have:

(v) We have:

(vi) We have:

(vii) We have:

(viii) We have:

(ix) We have:

#### Page No 94:

We shall use the identity (a+b)2 =a2 +b2 +2ab.

(i)
${\left(54\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(50+4{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(50\right)}^{2}+2×50×4+{\left(4\right)}^{2}\phantom{\rule{0ex}{0ex}}=2500+400+16\phantom{\rule{0ex}{0ex}}=2916$

(ii)
${\left(82\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(80+2{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(80\right)}^{2}+2×80×2+{\left(2\right)}^{2}\phantom{\rule{0ex}{0ex}}=6400+320+4\phantom{\rule{0ex}{0ex}}=6724$

(iii)
${\left(103\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(100+3{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(100\right)}^{2}+2×100×3+{\left(3\right)}^{2}\phantom{\rule{0ex}{0ex}}=10000+600+9\phantom{\rule{0ex}{0ex}}=10609$

(iv)
${\left(704\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(700+4{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(700\right)}^{2}+2×700×4+{\left(4\right)}^{2}\phantom{\rule{0ex}{0ex}}=490000+5600+16\phantom{\rule{0ex}{0ex}}=495616$

#### Page No 94:

We shall use the identity (a-b)2 = a2 +b2 -2ab.

(i)
${\left(69\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(70-1{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(70\right)}^{2}-2×70×1+1\phantom{\rule{0ex}{0ex}}=4900-140+1\phantom{\rule{0ex}{0ex}}=4761\phantom{\rule{0ex}{0ex}}$

(ii)
${\left(78\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(80-2{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(80\right)}^{2}-2×80×2+4\phantom{\rule{0ex}{0ex}}=6400-320+4\phantom{\rule{0ex}{0ex}}=6084\phantom{\rule{0ex}{0ex}}$

(iii)
${\left(197\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(200-3{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(200\right)}^{2}-2×200×3+9\phantom{\rule{0ex}{0ex}}=40000-1200+9\phantom{\rule{0ex}{0ex}}=38809\phantom{\rule{0ex}{0ex}}$

(iv)
${\left(999\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(1000-1{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(1000\right)}^{2}-2×1000×1+1\phantom{\rule{0ex}{0ex}}=1000000-2000+1\phantom{\rule{0ex}{0ex}}=998001\phantom{\rule{0ex}{0ex}}$

#### Page No 94:

We shall use the identity (a-b) (a+b)=a2 - b2.

(i)
$\left(82{\right)}^{2}-\left(18{\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(82-18\right)\left(82+18\right)\phantom{\rule{0ex}{0ex}}=\left(64\right)\left(100\right)\phantom{\rule{0ex}{0ex}}=6400$

(ii)
$\left(128{\right)}^{2}-\left(72{\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(128-72\right)\left(128+72\right)\phantom{\rule{0ex}{0ex}}=\left(56\right)\left(200\right)\phantom{\rule{0ex}{0ex}}=11200$

(iii)
$197×203\phantom{\rule{0ex}{0ex}}=\left(200-3\right)\left(200+3\right)\phantom{\rule{0ex}{0ex}}={\left(200\right)}^{2}-{\left(3\right)}^{2}\phantom{\rule{0ex}{0ex}}=40000-9\phantom{\rule{0ex}{0ex}}=39991$

(iv)
$\frac{198×198-102×102}{96}\phantom{\rule{0ex}{0ex}}=\frac{{\left(198\right)}^{2}-{\left(102\right)}^{2}}{96}\phantom{\rule{0ex}{0ex}}=\frac{\left(198-102\right)\left(198+102\right)}{96}\phantom{\rule{0ex}{0ex}}=\frac{\left(96\right)\left(300\right)}{96}\phantom{\rule{0ex}{0ex}}=300$

(v)
$\left(14.7×15.3\right)\phantom{\rule{0ex}{0ex}}=\left(15-0.3\right)×\left(15+0.3\right)\phantom{\rule{0ex}{0ex}}=\left(15{\right)}^{2}-\left(0.3{\right)}^{2}\phantom{\rule{0ex}{0ex}}=225-0.09\phantom{\rule{0ex}{0ex}}=224.91$

(vi)
$\left(8.63{\right)}^{2}-\left(1.37{\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(8.63-1.37\right)\left(8.63+1.37\right)\phantom{\rule{0ex}{0ex}}=\left(7.26\right)\left(10\right)\phantom{\rule{0ex}{0ex}}=72.6$

#### Page No 94:

Therefore, the value of the expression (9x2 + 24x + 16), when x = 12, is 1600.

#### Page No 94:

Therefore, the value of the expression (64x2 + 81y2 + 144xy), when x = 11 and y = $\frac{4}{3}$, is 10000.y=43

#### Page No 94:

Therefore, the value of  x2+$\frac{1}{{x}^{2}}$ is 14.

Therefore, the value of x4 + $\frac{1}{{x}^{4}}$ is 194.

(c) (−6a + 17b)

#### Page No 95:

(d) (3p2 + 5q − 9r3 +7)

(d) x2 + 2x − 15

#### Page No 95:

(b) (6x2 + 7x − 3)

#### Page No 95:

(c) (x2 + 8x + 16)

#### Page No 95:

(d) (x2 − 12x + 36)

(b) (4x2 − 25)

(c) −4ab2

(b) (2x + 1)

(a) (x − 2)

(c) (a4 − 1)

#### Page No 95:

a) $\left(\frac{1}{{x}^{2}}-\frac{1}{{y}^{2}}\right)$

(1x21y2)

(c) 23

(b) 38

#### Page No 95:

(c) 6400

[using the identity (a-b)(a+b)=a2 -b2]

#### Page No 95:

(a) 39991

$\left(197\right)×\left(203\right)\phantom{\rule{0ex}{0ex}}⇒\left(200-3\right)\left(200+3\right)\phantom{\rule{0ex}{0ex}}⇒{\left(200\right)}^{2}-{\left(3\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒40000-9\phantom{\rule{0ex}{0ex}}⇒39991$                   [using the identity (a+b) (a-b) = a2  -b2]

(b) 116