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Chapter 3

Expansions — Exercise 3.1

Class - 9 ML Aggarwal Understanding ICSE Mathematics

Exercise 3.1

Question 1(i)

By using standard formulae, expand the following:

(2x + 7y)2

Answer

(2x + 7y)2 = (2x)2 + 2(2x)(7y) + (7y)2

= 4x2 + 28xy + 49y2

Question 1(ii)

By using standard formulae, expand the following:

(12x+23y)2\Big(\dfrac{1}{2}x + \dfrac{2}{3}y\Big)^2

Answer

(12x+23y)2=(12x)2+2(12x)(23y)+(23y)2=14x2+23xy+49y2\Big(\dfrac{1}{2}x + \dfrac{2}{3}y\Big)^2 = \Big(\dfrac{1}{2}x\Big)^2 + 2\Big(\dfrac{1}{2}x\Big)\Big(\dfrac{2}{3}y\Big) + \Big(\dfrac{2}{3}y\Big)^2 \\[1em] = \dfrac{1}{4}x^2 + \dfrac{2}{3}xy + \dfrac{4}{9}y^2

Question 2(i)

By using standard formulae, expand the following:

(3x+12x)2\Big(3x + \dfrac{1}{2x} \Big)^2

Answer

(3x+12x)2=(3x)2+2(3x)(12x)+(12x)2=9x2+3+(14x2)\Big(3x + \dfrac{1}{2x} \Big)^2 = (3x)^2 + 2 (3x)\Big(\dfrac{1}{2x}\Big) + \Big(\dfrac{1}{2x}\Big)^2 \\[1em] = 9x^2 + 3 + \Big(\dfrac{1}{4x^2}\Big)

Question 2(ii)

By using standard formulae, expand the following:

(3x2y + 5z)2

Answer

(3x2y + 5z)2 = (3x2y)2 + 2 (3x2y)(5z) + (5z)2

= 9x4y2 + 30x2yz + 25z2

Question 3(i)

By using standard formulae, expand the following:

(3x−12x)2\Big(3x - \dfrac{1}{2x} \Big)^2

Answer

(3x−12x)2=(3x)2−2(3x)(12x)+(12x)2=9x2−3+(14x2)\Big(3x - \dfrac{1}{2x} \Big)^2 = (3x)^2 - 2 (3x)\Big(\dfrac{1}{2x}\Big) + \Big(\dfrac{1}{2x}\Big)^2 \\[1em] = 9x^2 - 3 + \Big(\dfrac{1}{4x^2}\Big)

Question 3(ii)

By using standard formulae, expand the following:

(12x−32y)2\Big(\dfrac{1}{2}x - \dfrac{3}{2}y\Big)^2

Answer

(12x−32y)2=(12x)2−2(12x)(32y)+(32y)2=14x2−32xy+94y2\Big(\dfrac{1}{2}x - \dfrac{3}{2}y\Big)^2 = \Big(\dfrac{1}{2}x\Big)^2 - 2\Big(\dfrac{1}{2}x\Big)\Big(\dfrac{3}{2}y\Big) + \Big(\dfrac{3}{2}y\Big)^2 \\[1em] = \dfrac{1}{4}x^2 - \dfrac{3}{2}xy + \dfrac{9}{4}y^2

Question 4(i)

By using standard formulae, expand the following:

(x+3)(x+5)

Answer

We know that (x+a)(x+b) = x2 + (a + b)x + ab

∴ (x+3)(x+5) = x2 + (3 + 5)x + (3)(5) = x2 + 8x + 15

Question 4(ii)

By using standard formulae, expand the following:

(x+3)(x-5)

Answer

We know that (x+a)(x-b) = x2 + (a - b)x - ab

∴ (x+3)(x-5) = x2 + (3 - 5)x - (3)(5) = x2 - 2x - 15

Question 4(iii)

By using standard formulae, expand the following:

(x-7)(x+9)

Answer

We know that (x-a)(x+b) = x2 - (a - b)x - ab

∴ (x-7)(x+9) = x2 - (7 - 9)x - (7)(9) = x2 + 2x - 63

Question 4(iv)

By using standard formulae, expand the following:

(x-2y)(x-3y)

Answer

We know that (x-a)(x-b) = x2 - (a + b)x + ab

∴ (x-2y)(x-3y) = x2 - (2y + 3y)x + (2y)(3y)

= x2 - 5xy + 6y2

Question 5(i)

By using standard formulae, expand the following:

(x - 2y - z)2

Answer

(x - 2y - z)2 = [(x) + (-2y) + (-z)]2

= (x)2 + (-2y)2 + (-z)2 + 2 [(x)(-2y) + (-2y)(-z) + (-z)(x)]

= x2 + 4y2 + z2 + 2[-2xy + 2yz - xz]

= x2 + 4y2 + z2 - 4xy + 4yz - 2xz

Question 5(ii)

By using standard formulae, expand the following:

(2x - 3y + 4z)2

Answer

(2x - 3y + 4z)2 = [(2x) + (-3y) + (4z)]2

= (2x)2 + (-3y)2 + (4z)2 + 2 [(2x)(-3y) + (-3y)(4z) + (4z)(2x)]

= 4x2 + 9y2 + 16z2 + 2[-6xy - 12yz + 8xz]

= 4x2 + 9y2 + 16z2 - 12xy - 24yz + 16xz

Question 6(i)

By using standard formulae, expand the following:

(2x+3x−1)2\Big(2x + \dfrac{3}{x} - 1\Big)^2

Answer

(2x+3x−1)2=[(2x)+(3x)+(−1)]2=(2x)2+(3x)2+(−1)2+2[(2x)(3x)+(3x)(−1)+(−1)(2x)]=4x2+9x2+1+2[6−3x−2x]=4x2+9x2+1+12−6x−4x=4x2+9x2+13−6x−4x\Big(2x + \dfrac{3}{x} - 1\Big)^2 = \Big[\Big(2x\Big) + \Big(\dfrac{3}{x}\Big) + \Big(-1\Big)\Big]^2 \\[1em] = (2x)^2 + \Big(\dfrac{3}{x}\Big)^2 + (-1)^2 + 2 \Big[\Big(2x\Big)\Big(\dfrac{3}{x}\Big) + \Big(\dfrac{3}{x}\Big)\Big(-1\Big)+\Big(-1\Big)\Big(2x\Big) \Big] \\[1em] = 4x^2 + \dfrac{9}{x^2} + 1 + 2\Big[6 - \dfrac{3}{x} - 2x \Big] \\[1em] = 4x^2 + \dfrac{9}{x^2} + 1 + 12 - \dfrac{6}{x} - 4x \\[1em] = 4x^2 + \dfrac{9}{x^2} + 13 - \dfrac{6}{x} - 4x \\[1em]

Question 6(ii)

By using standard formulae, expand the following:

(23x−32x−1)2\Big(\dfrac{2}{3}x - \dfrac{3}{2x} - 1\Big)^2

Answer

(23x−32x−1)2=[23x+(−32x)+(−1)]2=(23x)2+(−32x)2+(−1)2+2[(23x)(−32x)+(−32x)(−1)+(−1)(23x)]=49x2+94x2+1+2[−1+32x−23x]=49x2+94x2+1−2+3x−43x=49x2+94x2−1+3x−43x\Big(\dfrac{2}{3}x - \dfrac{3}{2x} - 1\Big)^2 = \Big[\dfrac{2}{3}x + \Big(-\dfrac{3}{2x}\Big) + \Big(-1\Big)\Big]^2 \\[1em] = \Big(\dfrac{2}{3}x\Big)^2 + \Big(-\dfrac{3}{2x}\Big)^2 + (-1)^2 + 2 \Big[\Big(\dfrac{2}{3}x\Big)\Big(-\dfrac{3}{2x}\Big) + \Big(-\dfrac{3}{2x}\Big)\Big(-1\Big)+\Big(-1\Big)\Big(\dfrac{2}{3}x\Big) \Big] \\[1em] = \dfrac{4}{9}x^2 + \dfrac{9}{4x^2} + 1 + 2\Big[-1 + \dfrac{3}{2x} - \dfrac{2}{3}x \Big] \\[1em] = \dfrac{4}{9}x^2 + \dfrac{9}{4x^2} + 1 - 2 + \dfrac{3}{x} - \dfrac{4}{3}x \\[1em] = \dfrac{4}{9}x^2 + \dfrac{9}{4x^2} -1 + \dfrac{3}{x} - \dfrac{4}{3}x \\[1em]

Question 7(i)

By using standard formulae, expand the following:

(x+2)3

Answer

(x+2)3 = x3 + (2)3 + 3(x)(2)(x+2)

= x3 + 8 + 6x2 + 12x

Question 7(ii)

By using standard formulae, expand the following:

(2a+b)3

Answer

(2a+b)3 = (2a)3 + (b)3 + 3(2a)(b)(2a+b)

= 8a3 + b3 + 12a2b + 6ab2

Question 8(i)

By using standard formulae, expand the following:

(3x+1x)3\Big(3x + \dfrac{1}{x}\Big)^3

Answer

(3x+1x)3=(3x)3+(1x)3+3(3x)(1x)(3x+1x)=27x3+1x3+27x+9x\Big(3x + \dfrac{1}{x}\Big)^3 = (3x)^3 + \Big(\dfrac{1}{x}\Big)^3 + 3(3x)\Big(\dfrac{1}{x}\Big)\Big(3x + \dfrac{1}{x}\Big) \\[1em] = 27x^3 + \dfrac{1}{x^3} + 27x + \dfrac{9}{x}

Question 8(ii)

By using standard formulae, expand the following:

(2x-1)3

Answer

(2x-1)3 = (2x)3 - (1)3 - 3(2x)(1)(2x-1)

= 8x3 - 1 -12x2 + 6x

Question 9(i)

By using standard formulae, expand the following:

(5x-3y)3

Answer

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

= 125x3 - 27y3 - 225x2y + 135xy2

Question 9(ii)

By using standard formulae, expand the following:

(2x−13y)3\Big(2x - \dfrac{1}{3y}\Big)^3

Answer

(2x−13y)3=(2x)3−(13y)3−3(2x)(13y)(2x−13y)=8x3−127y3−4x2y+2x3y2\Big(2x - \dfrac{1}{3y}\Big)^3 = (2x)^3 - \Big(\dfrac{1}{3y}\Big)^3 - 3(2x)\Big(\dfrac{1}{3y}\Big)\Big(2x - \dfrac{1}{3y}\Big) \\[1em] = 8x^3 - \dfrac{1}{27y^3} - \dfrac{4x^2}{y} + \dfrac{2x}{3y^2}

Question 10(i)

Simplify the following:

(a+1a)2+(a−1a)2\Big(a + \dfrac{1}{a}\Big)^2 + \Big(a - \dfrac{1}{a}\Big)^2

Answer

(a+1a)2+(a−1a)2=[(a)2+2(a)(1a)+(1a)2]+[(a)2−2(a)(1a)+(1a)2]=[a2+2+1a2]+[a2−2+1a2]=2a2+2a2=2(a2+1a2)\Big(a + \dfrac{1}{a}\Big)^2 + \Big(a - \dfrac{1}{a}\Big)^2 = \Big[\Big(a\Big)^2 + 2\Big(a\Big)\Big(\dfrac{1}{a}\Big) + \Big(\dfrac{1}{a}\Big)^2 \Big] + \Big[\Big(a\Big)^2 - 2\Big(a\Big)\Big(\dfrac{1}{a}\Big) + \Big(\dfrac{1}{a}\Big)^2\Big] \\[1em] = \Big[a^2 + 2 + \dfrac{1}{a^2} \Big] + \Big[a^2 - 2 + \dfrac{1}{a^2} \Big] \\[1em] = 2a^2 + \dfrac{2}{a^2} \\[1em] = 2\Big(a^2 + \dfrac{1}{a^2}\Big)

Question 10(ii)

Simplify the following:

(a+1a)2−(a−1a)2\Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2

Answer

(a+1a)2−(a−1a)2=[(a)2+2(a)(1a)+(1a)2]−[(a)2−2(a)(1a)+(1a)2]=[a2+2+1a2]−[a2−2+1a2]=a2+2+1a2−a2+2−1a2=4\Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 = \Big[\Big(a\Big)^2 + 2\Big(a\Big)\Big(\dfrac{1}{a}\Big) + \Big(\dfrac{1}{a}\Big)^2 \Big] - \Big[\Big(a\Big)^2 - 2\Big(a\Big)\Big(\dfrac{1}{a}\Big) + \Big(\dfrac{1}{a}\Big)^2\Big] \\[1em] = \Big[a^2 + 2 + \dfrac{1}{a^2} \Big] - \Big[a^2 - 2 + \dfrac{1}{a^2} \Big] \\[1em] = a^2 + 2 + \dfrac{1}{a^2} - a^2 + 2 - \dfrac{1}{a^2} = 4

Question 11(i)

Simplify the following:

(3x-1)2 - (3x-2)(3x+1)

Answer

(3x-1)2 - (3x-2)(3x+1) = [(3x-1)2] - [(3x-2)(3x+1)]

= [(3x)2 - 2(3x)(1) + (1)2] - [(3x)2 + 3x - 6x - 2]

= (9x2 - 6x + 1) - (9x2 - 3x - 2)

= 9x2 - 6x + 1 - 9x2 + 3x + 2

= -3x + 3

= 3 - 3x

Question 11(ii)

Simplify the following:

(4x+3y)2 - (4x-3y)2 - 48xy

Answer

(4x+3y)2 - (4x-3y)2 - 48xy = [(4x)2 + 2(4x)(3y) + (3y)2] - [(4x)2 - 2(4x)(3y) + (3y)2] - 48xy

= [16x2 + 24xy + 9y2] - [16x2 - 24xy + 9y2] - 48xy

= 16x2 + 24xy + 9y2 - 16x2 + 24xy - 9y2 - 48xy

= 48xy - 48xy = 0

Question 12(i)

Simplify the following:

(7p+9q)(7p-9q)

Answer

(7p+9q)(7p-9q) = (7p)2 - (9q)2

= 49p2 - 81q2

Question 12(ii)

Simplify the following:

(2x−3x)(2x+3x)\Big(2x - \dfrac{3}{x}\Big)\Big(2x + \dfrac{3}{x}\Big)

Answer

(2x−3x)(2x+3x)=(2x)2−(3x)2=4x2−9x2\Big(2x - \dfrac{3}{x}\Big)\Big(2x + \dfrac{3}{x}\Big) = \Big(2x\Big)^2 - \Big(\dfrac{3}{x}\Big)^2 = 4x^2 - \dfrac{9}{x^2}

Question 13(i)

Simplify the following:

(2x - y + 3)(2x - y - 3)

Answer

(2x - y + 3)(2x - y - 3)

Let (2x - y) = a

Then, the given expression = (a+3)(a-3) = (a)2 - (3)2

= a2 - 9 = (2x-y)2 - 9 = (2x)2 - 2(2x)(y) + y2 - 9

= 4x2 - 4xy + y2 - 9

Question 13(ii)

Simplify the following:

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

Answer

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

Let (3x-5)=a

Then, the given expression = (a+y)(a-y)

= a2 - y2 = (3x-5)2 - y2

= [(3x)2 - 2(3x)(5) + (5)2] - y2

= 9x2 - 30x + 25 - y2

Question 14(i)

Simplify the following:

(x+2x−3)(x−2x−3)\Big(x + \dfrac{2}{x} - 3\Big)\Big(x - \dfrac{2}{x} - 3\Big)

Answer

(x+2x−3)(x−2x−3)\Big(x + \dfrac{2}{x} - 3\Big)\Big(x - \dfrac{2}{x} - 3\Big) \\[1em]

Let (x - 3) = a

Then, the given expression = (a+2x)(a−2x)=(a)2−(2x)2=a2−4x2=(x−3)2−4x2=x2−2(x)(3)+32−4x2x2−6x+9−4x2\Big(a+\dfrac{2}{x}\Big)\Big(a-\dfrac{2}{x}\Big) = \Big(a\Big)^2 - \Big(\dfrac{2}{x}\Big)^2 \\[1em] = a^2 - \dfrac{4}{x^2} = (x-3)^2 - \dfrac{4}{x^2} = x^2 -2(x)(3) + 3^2 - \dfrac{4}{x^2} \\[1em] x^2 - 6x + 9 - \dfrac{4}{x^2}

Question 14(ii)

Simplify the following:

(5-2x) (5+2x) (25+4x2)

Answer

(5-2x)(5+2x)(25+4x2) = [(5)2 - (2x)2] (25+4x2)

= (25-4x2)(25+4x2)

= (25)2 - (4x2)2 = 625 - 16x4

Question 15(i)

Simplify the following:

(x+2y+3)(x+2y+7)

Answer

(x+2y+3)(x+2y+7)

Let (x+2y)=z

Then the given expression = (z+3)(z+7)

We know that (x+a)(x+b) = x2 + (a + b)x + ab

∴ (z+3)(z+7) = z2 + (3+7)z + (3)(7)

= z2 + 10z + 21

= (x+2y)2 + 10(x+2y) + 21

= x2 + 2(x)(2y) + (2y)2 + 10x + 20y + 21

= x2 + 4xy + 4y2 + 10x + 20y + 21

= x2 + 10x + 4y2 + 20y + 4xy + 21

Question 15(ii)

Simplify the following:

(2x+y+5)(2x+y-9)

Answer

(2x+y+5)(2x+y-9)

Let (2x+y)=z

Then the given expression = (z+5)(z-9)

We know that (x+a)(x-b) = x2 + (a - b)x - ab

∴ (z+5)(z-9) = z2 + (5-9)z - (5)(9)

= z2 - 4z - 45

= (2x+y)2 - 4(2x+y) - 45

= (2x)2 + 2(2x)(y) + (y)2 - 8x - 4y - 45

= 4x2 + 4xy + y2 - 8x - 4y - 45

= 4x2 - 8x + y2 - 4y + 4xy - 45

Question 15(iii)

Simplify the following:

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

Answer

Let (x - 2y) = z

Then the given expression = (z - 5)(z + 3)

We know that (x - a)(x + b) = x2 - (a - b)x - ab

∴ (z - 5)(z + 3) = z2 - (5 - 3)z - (5)(3)

= z2 - 2z - 15

= (x - 2y)2 - 2(x - 2y) - 15

= (x)2 - 2(x)(2y) + (2y)2 - 2x + 4y - 15

= x2 - 4xy + 4y2 - 2x + 4y - 15

Question 15(iv)

Simplify the following:

(3x - 4y - 2)(3x - 4y - 6)

Answer

Let (3x - 4y) = z

Then the given expression = (z - 2)(z - 6)

We know that (x - a)(x - b) = x2 - (a + b)x + ab

∴ (z - 2)(z - 6) = z2 - (2 + 6)z + (2)(6)

= z2 - 8z + 12

= (3x - 4y)2 - 8(3x - 4y) + 12

= (3x)2 - 2(3x)(4y) + (4y)2 - 24x + 32y + 12

= 9x2 - 24xy + 16y2 - 24x + 32y + 12

Question 16(i)

Simplify the following:

(2p + 3q)(4p2 - 6pq + 9q2)

Answer

(2p + 3q)(4p2 - 6pq + 9q2) = (2p + 3q)[(2p)2 - (2p)(3q) + (3q)2]

We know that (a + b)(a2 - ab + b2) = a3 + b3

∴ (2p + 3q)[(2p)2 - (2p)(3q) + (3q)2] = (2p)3 + (3q)3

= 8p3 + 27q3

Question 16(ii)

Simplify the following:

(x+1x)(x2−1+1x2)\Big(x + \dfrac{1}{x}\Big)\Big(x^2 - 1 + \dfrac{1}{x^2}\Big)

Answer

We know that (a + b)(a2 - ab + b2) = a3 + b3

∴ Given Expression = (x+1x)(x2−1+1x2)\Big(x + \dfrac{1}{x}\Big)\Big(x^2 - 1 + \dfrac{1}{x^2}\Big)

=x3+1x3= x^3 + \dfrac{1}{x^3}

Question 17(i)

Simplify the following:

(3p - 4q)(9p2 + 12pq + 16q2)

Answer

(3p - 4q)(9p2 +12pq + 16q2) = (3p - 4q)[(3p)2 + (3p)(4q) + (4q)2]

We know that (a - b)(a2 + ab + b2) = a3 - b3

∴ (3p - 4q)[(3p)2 + (3p)(4q) + (4q)2] = (3p)3 - (4q)3

= 27p3 - 64q3

Question 17(ii)

Simplify the following:

(x−3x)(x2+3+9x2)\Big(x - \dfrac{3}{x}\Big)\Big(x^2 + 3 + \dfrac{9}{x^2}\Big)

Answer

We know that (a - b)(a2 + ab + b2) = a3 - b3

∴ Given Expression = (x−3x)(x2+3+9x2)\Big(x - \dfrac{3}{x}\Big)\Big(x^2 + 3 + \dfrac{9}{x^2}\Big)

=x3−(3x)3=x3−27x3= x^3 - \Big(\dfrac{3}{x}\Big)^3 \\[1em] = x^3 - \dfrac{27}{x^3}

Question 18

Simplify the following:

(2x + 3y + 4z)(4x2 + 9y2 + 16z2 - 6xy - 12yz - 8zx)

Answer

We know that (a + b + c)(a2 + b2 + c2 - ab - bc - ca) = a3 + b3 + c3 - 3abc

∴ The Given Expression = (2x + 3y + 4z)[(2x)2 + (3y)2 + (4z)2 - (2x)(3y) - (3y)(4z) - (4z)(2x)]

= (2x)3 + (3y)3 + (4z)3 - 3(2x)(3y)(4z)

= 8x3 + 27y3 + 64z3 - 72xyz

Question 19(i)

Find the product of the following:

(x + 1)(x + 2)(x + 3)

Answer

We know that (x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc

∴ (x + 1)(x + 2)(x + 3) = x3 + (1 + 2 + 3)x2 + [(1)(2) + (2)(3) + (3)(1)]x + (1)(2)(3)

= x3 + 6x2 + (2 + 6 + 3)x + 6

= x3 + 6x2 + 11x + 6

Question 19(ii)

Find the product of the following:

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

Answer

We know that (x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc

∴ (x - 2)(x - 3)(x + 4) = x3 + [(-2) + (-3) + 4]x2 + [(-2)(-3) + (-3)(4) + (4)(-2)]x + (-2)(-3)(4)

= x3 - x2 + (6 -12 - 8)x + 24

= x3 - x2 - 14x + 24

Question 20

Find the coefficient of x2 and x in the product of (x - 3)(x + 7)(x - 4)

Answer

We know that (x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc

Comparing (x - 3)(x + 7)(x - 4) with (x + a)(x + b)(x + c).
Here a = -3, b = 7, c = -4.

∴ Coefficient of x2 = a + b + c = (-3) + 7 + (-4) = 0 and
    coefficient of x = ab + bc + ca

    = (-3) x 7 + 7 x (-4) + (-4) x (-3)

    = -21 - 28 + 12 = -37

Question 21

If a2 + 4a + x = (a + 2)2, find the value of x

Answer

Given,
    a2 + 4a + x = (a + 2)2

⇒ a2 + 4a + x = a2 + 2(a)(2) + 22

⇒ a2 - a2 +4a - 4a + x = 4

⇒ x = 4

Question 22(i)

Use (a+b)2 = a2 + 2ab + b2 to evaluate the following:

(101)2

Answer

(101)2 = (100 + 1)2

Using (a + b)2 = a2 + 2ab + b2,

(100 + 1)2 = (100)2 + 2(100)(1) + (1)2

= 10000 + 200 + 1 = 10201

Question 22(ii)

Use (a+b)2 = a2 + 2ab + b2 to evaluate the following:

(1003)2

Answer

(1003)2 = (1000 + 3)2

Using (a + b)2 = a2 + 2ab + b2,

(1000 + 3)2 = (1000)2 + 2(1000)(3) + (3)2

= 1000000 + 6000 + 9 = 1006009

Question 22(iii)

Use (a+b)2 = a2 + 2ab + b2 to evaluate the following:

(10.2)2

Answer

(10.2)2 = (10 + 0.2)2

Using (a + b)2 = a2 + 2ab + b2,

(10 + 0.2)2 = (10)2 + 2(10)(0.2) + (0.2)2

= 100 + 4 + 0.04 = 104.04

Question 23(i)

Use (a - b)2 = a2 - 2ab + b2 to evaluate the following:

(99)2

Answer

(99)2 = (100 - 1)2

Using (a - b)2 = a2 - 2ab + b2,

(100 - 1)2 = (100)2 - 2(100)(1) + (1)2

= 10000 - 200 + 1 = 9801

Question 23(ii)

Use (a - b)2 = a2 - 2ab + b2 to evaluate the following:

(997)2

Answer

(997)2 = (1000 - 3)2

Using (a - b)2 = a2 - 2ab + b2,

(1000 - 3)2 = (1000)2 - 2(1000)(3) + (3)2

= 1000000 - 6000 + 9 = 994009

Question 23(iii)

Use (a - b)2 = a2 - 2ab + b2 to evaluate the following:

(9.8)2

Answer

(9.8)2 = (10 - 0.2)2

Using (a - b)2 = a2 - 2ab + b2,

(10 - 0.2)2 = (10)2 - 2(10)(0.2) + (0.2)2

= 100 - 4 + 0.04 = 96.04

Question 24(i)

By using suitable identities, evaluate the following:

(103)3

Answer

(103)3 = (100 + 3)3

Using (a + b)3 = a3 + b3 + 3(a)(b)(a + b),

(100 + 3)3 = (100)3 + (3)3 + 3(100)(3)(100 + 3)

= 1000000 + 27 + 900(103)

= 1000000 + 27 + 92700

= 1092727

Question 24(ii)

By using suitable identities, evaluate the following:

(99)3

Answer

(99)3 = (100 - 1)3

Using (a - b)3 = a3 - b3 - 3(a)(b)(a - b),

(100 - 1)3 = (100)3 - (1)3 - 3(100)(1)(100 - 1)

= 1000000 - 1 - 300(99)

= 1000000 - 1 - 29700

= 970299

Question 24(iii)

By using suitable identities, evaluate the following:

(10.1)3

Answer

(10.1)3 = (10 + 0.1)3

Using (a + b)3 = a3 + b3 + 3(a)(b)(a + b),

(10 + 0.1)3 = (10)3 + (0.1)3 + 3(10)(0.1)(10.1)

= 1000 + 0.001 + 30.3

= 1030.301

Question 25

If 2a - b + c = 0, prove 4a2 - b2 + c2 + 4ac = 0

Answer

Given,
     2a - b + c = 0
⇒ (2a + c) = b

On squaring both sides,

    (2a + c)2 = b2

⇒ 4a2 + c2 + 2(2a)(c) = b2

⇒ 4a2 - b2 + c2 + 4ac = 0

Hence proved.

Question 26

If a + b + 2c = 0, prove that a3 + b3 + 8c3 = 6abc

Answer

We know that if a + b + c = 0 then a3 + b3 + c3 = 3abc

Given,

a + b + 2c = 0

∴ (a)3 + (b)3 + (2c)3 = 3(a)(b)(2c)

⇒ a3 + b3 + 8c3 = 6abc

Hence Proved.

Question 27

If x + 2y - 3 = 0, then find the value of x3 + 8y3 + 6x2y + 12xy2 - 125.

Answer

Given,

x + 2y - 3 = 0

⇒ x + 2y = 3

Simplifying,

⇒ x3 + 8y3 + 6x2y + 12xy2 - 125

⇒ x3 + (2y)3 + 6xy(x + 2y) - 125

⇒ x3 + (2y)3 + 3.x.2y.(x + 2y) - 125

⇒ (x + 2y)3 - 125

⇒ 33 - 125

⇒ 27 - 125

⇒ -98.

Hence, the value of x3 + 8y3 + 6x2y + 12xy2 - 125 = -98.

Question 28

If a + b + c = 0, then find the value of a2bc+b2ca+c2ab\dfrac{a^2}{bc} + \dfrac{b^2}{ca} + \dfrac{c^2}{ab}.

Answer

We know if a + b + c = 0, a3 + b3 + c3 = 3abc.

On dividing each side by abc,
⇒a3abc+b3abc+c3abc=3abcabc⇒a2bc+b2ca+c2ab=3\phantom{\Rightarrow} \dfrac{a^3}{abc} + \dfrac{b^3}{abc} + \dfrac{c^3}{abc} = \dfrac{3abc}{abc} \\[1em] \Rightarrow \dfrac{a^2}{bc} + \dfrac{b^2}{ca} + \dfrac{c^2}{ab} = 3

∴ Value of a2bc+b2ca+c2ab=3\bold{\dfrac{a^2}{bc} + \dfrac{b^2}{ca} + \dfrac{c^2}{ab} = 3}

Question 29

If x + y = 4, find the value of x3 + y3 + 12xy - 64

Answer

Using (x + y)3 = x3 + y3 + 3(x)(y)(x + y)

Putting x + y = 4 in above:

    x3 + y3 + 3xy(4) = (4)3

⇒ x3 + y3 + 12xy = 64

⇒ x3 + y3 + 12xy - 64 = 0

∴ Value of x3 + y3 + 12xy - 64 = 0

Question 30

Without actually calculating the cubes, find the values of:

(i) (27)3 + (-17)3 + (-10)3

(ii) (-28)3 + (15)3 + (13)3

Answer

(i) Let a = 27, b = -17, c = -10

Then,

a + b + c = 37 - 17 - 10 = 0

Thus, a3 + b3 + c3 = 3abc

∴ (27)3 + (-17)3 + (-10)3 = 3(27)(-17)(-10) = 13770

(ii) Let a = -28, b = 15, c = 13

Then,

a + b + c = -28 + 15 + 13 = 0

Thus, a3 + b3 + c3 = 3abc

∴ (-28)3 + (15)3 + (13)3 = 3(-28)(15)(13) = -16380

Question 31

Using suitable identity, find the value of :

86×86×86+14×14×1486×86−86×14+14×14\dfrac{86 \times 86 \times 86 + 14 \times 14 \times 14}{86 \times 86 - 86 \times 14 + 14 \times 14}

Answer

Let x = 86 and y = 14.

Hence, above equation can be written as,

⇒x3+y3x2−xy+y2=(x+y)(x2−xy+y2)(x2−xy+y2)=x+y=86+14=100.\Rightarrow \dfrac{x^3 + y^3}{x^2 - xy + y^2} \\[1em] = \dfrac{(x + y)(x^2 - xy + y^2)}{(x^2 - xy + y^2)} \\[1em] = x + y \\[1em] = 86 + 14 \\[1em] = 100.

Hence, value of 86×86×86+14×14×1486×86−86×14+14×14\dfrac{86 \times 86 \times 86 + 14 \times 14 \times 14}{86 \times 86 - 86 \times 14 + 14 \times 14} = 100.

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