ML Aggarwal Solution Class 9 Chapter 4 Factorisation Exercise 4.5

 Exercise 4.5

Question 1

(i) 8x³+y³

⇒(2x)³+y³

∴It is in the form of a³+b³=(a+b)(a²-ab+b²)

∴there a=2x ; b=7

∴⇒(2x)³+y³=(2x+y)[(2x)²-2x.y+y²]

=(2x+y)(4x²-2xy+y²)

(ii) 64x³-125y³

⇒(4a)³-(5y)³

⇒It is in the form of a³-b³=(a-b)(a²+ab+b²)

⇒here a=4x ; b=5y

∴⇒(4x-5y)[(4x)²+4x.5y+(5y)²]

⇒(4a-5y)(16x²+20xy+25y²)

Question 2

Sol :

(i) 64x³+1

⇒(4x)³+1³

⇒It is in the form of a³+b³=(a+b)(a²-ab+b²)

Here a=4x ; b=1

∴⇒(4x+1)[(4x)²-4x.1+1²]

⇒(4x+1)(16x²-4x+1)

(ii) 7a³+56b³

⇒7(a³+8b³)

⇒7(a³+(2b)³)

It is in the form of a³+b³=(a+b)(a²-ab+b²)

∴Here a=a ; b=2b

∴⇒7(a+2b)(a²-a.2b+(2b)²)

⇒7(a+2b)(a²-2ab+4b²)

Question 3

(i) $\frac{x^6}{343}+\frac{343}{x^6}$

$\frac{x^6}{7^3}+\frac{7^3}{x^6}$

⇒It is in the form of a³+b³=(a+b)(a²-ab+b²)

∴Here $a=\frac{x^2}{7}$ ; $b=\frac{7}{x^2}$

⇒$\frac{(x^2)^3}{7^3}+\frac{7^3}{(x^2)^3}$

⇒$\left(\frac{x^2}{7}+\frac{7}{x^2}\right)\left[\left(\frac{x^2}{7}\right)^2-\frac{x^2}{7}\times \frac{7}{x^2}+\left(\frac{7}{x^2}\right)^2\right]$

⇒$\left(\frac{x^2}{7}+\frac{7}{x^2}\right)\left(\frac{x^4}{49}-1+\frac{49}{x^4}\right)$

(ii) $8x^3-\frac{1}{27y^3}$

⇒$(2x)^3-\frac{1}{(3y)}^3$

⇒It is in the form of a³+b³=(a+b)(a²-ab+b²)

Here a=2x ; $b=\frac{1}{3y}$

⇒$\left(2x-\frac{1}{3y}\right)\left[(2x)^2+2x.\frac{1}{3y}+(3y)^2\right]$

⇒$\left(2x-\frac{1}{3y}\right)\left(4x^2+\frac{2x}{3y}+9y^2\right)$

Question 4

(i) x²+x⁵

⇒x²(1+x³)

⇒x³(1³+x³)

It is in the form of a³+b³=(a+b)(a²-ab+b²)

Here a=1 ; b=x

∴x²(1+x)(1²-1.x+x²)

⇒x²(1+x)(1-x+x²)

(ii) 32x⁴-500

⇒4x(8x³-125)

⇒4x((2x)³-5³)

It is in the form of a³+b³=(a+b)(a²-ab+b²)

Here a=2x ; b=5

⇒4x(2x-5)((2x)²+2x.5+5²)

⇒4x(2x-5)(4x²+10x+25)

Question 5

(i) 27x³y³-8

⇒(3xy)³-2³

It is in the form of a³+b³=(a+b)(a²-ab+b²)

Here a=3xy ; b=2

⇒(3xy-2)[(3xy)²+3xy.2+2²]

⇒(3xy-2)[9x²y²+6xy+4]

(ii) 27(x+y)³+8(2x-y)³

⇒3³(x+y)³+2³(2x-y)³

⇒(3(x+y)³)+(2(x-y))³

It is in the form of a³+b³=(a+b)(a²-ab+b²)

⇒Here a=3(x+y) ; b=2(x-y)

⇒[3(x+y)+2(2x-y)][3²(x+y)²-3(x+y)(2x-y).2+2²(2x-y)²]

⇒(3x+3y+4x-2y)[9(x+y)²-6(x+y)(2x-y)+4(2x-y)²]

⇒(7x-y)[9(x²+y²+2xy)-6(2x²-xy+2xy-y²)+4(4x²+y²-4xy)]

⇒(7x-y)[9x²+9y²+18xy-12x²-6xy-6y²+16x²+4y²-16xy]

⇒(7x-y)[13x²-4xy+19y²]

Question 6

(i) a³+b³+a+b

⇒(a³+b³)+(a+b)

⇒(a+b)(a²-ab+b²)+(a+b)

⇒(a+b)(a²-ab+b²+1)

(ii) a³-b³-a+b

⇒(a³-b³)-(a-b)

⇒(a-b)(a²+ab+b²)-(a-b)

⇒(a-b)(a²+ab+b²-1)

Question 7

(i) x³+x+2

⇒x³+x+1+1

⇒(x³+1)+(x+1)

⇒(x+1)(x²-x+1)+(x+1)

⇒(x+1)(x²-x+1+1)

⇒(x+1)(x²-x+2)

(ii) a³-a-120

⇒a³-a-125+5

⇒a³-125-(a-5))

⇒(a³-5³)-(a-5)

⇒(a-5)(a²+5a+5²)-(a-5)

⇒(a-5)(a²+5a+25)-(a-5)

⇒(a-5)[a²+5a+25-1]

⇒(a-5)(a²+5a+24)

Question 8

(i) x³+6x²+12x+16

⇒x³+6x²+12x+8+18

⇒(x³+3.2.x²+3.2².x+2³)+8

It is in the form of a³+b³=(a+b)(a²-ab+b²)

∴Here a=x ; b=2

∴(x+2)³+8

⇒(x+2)³+2³

⇒(x+2+2)[(x+2)²-2.(x+2)+2²]

⇒(x+4)(x²+4+4x-2x-4+4)

⇒(x+4)(x²+2x+4)

(ii) a³-3a²b+3ab²-2b³

⇒a³-3a²b+3ab²-b³-b³

⇒(a-b)³-b³

It is in the form of a³+b³=(a+b)(a²-ab+b²)

Here a=a-b ; b=b

⇒(a-2b)((a-b)²+(a-b)b+b²)

⇒(a-2b)(a+b-2ab+ab-b²+b²)

⇒(a-2b)(a²+b²-ab)

Question 9

(i) 2a³+16b³-5a-10b

⇒a³+a³+8b³-5a-10b+8b³

⇒2(a³+(2b)³)-5(a+2b)

It is in the form of a³+b³=(a+b)(a²-ab+b²)

⇒2[(a+2b)(a²-2ab+(2b)a²)-5(a+2b)]

⇒2[(a+2b)(a²-2ab+4b²)]-5(a+2b)

⇒(a+2b)[2(a²-2ab+4b²)-5]

⇒(a+2b)[2a²-4ab+8b²-5]

(ii) $a^3-\frac{1}{a^3}-2a+\frac{2}{a}$

⇒a³-2a+a-a+$\frac{2}{a}+\frac{1}{a}-\frac{1}{a}-\frac{1}{a^3}$

⇒a³-3a+a+$\frac{3}{a}-\frac{1}{a}-\frac{1}{a^3}$

⇒a³-3a².$\frac{1}{a}+3.a.\frac{1}{a^2}-\frac{1}{a^3}+\left(a-\frac{1}{a}\right)$

It is in the form of a³+b³=(a+b)(a²-ab+b²)

∴Here a=a ; $b=\frac{1}{a}$

∴$\left(a-\frac{1}{a}\right)^3+\left(a-\frac{1}{a}\right)$

⇒$\left(a-\frac{1}{a}\right)\left[\left(a-\frac{1}{a}\right)^2+1\right]$

⇒$\left(a-\frac{1}{a}\right)\left[a^2+\frac{1}{a^2}-2.a.\frac{1}{a}+1\right]$

⇒$\left(a-\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}-2+1\right)$

⇒$\left(a-\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}-1\right)$

Question 10

(i) a⁶-b⁶

⇒(a²)³-(b²)³

It is in the form of a³+b³=(a+b)(a²-ab+b²)

Here a=a² ; b=b²

∴(a²-b²)[(a²)²+a²-b²+(b²)²]

⇒(a²-b²)(a⁴+a²b²+b⁴)

(ii) x⁶-1

⇒(x²)³-1³

⇒It is in the form of a³+b³=(a+b)(a²-ab+b²)

Here x²=a ; b=1

∴(x²-1)[(x²)²+x².1+1²]

⇒(x²-1)(x⁴+x²+1)

Question 11

(i) 64x⁶-729y⁶

⇒(2x)⁶-(3y)⁶

⇒[(2x)²]³-[(3y)²]³

⇒It is in the form of a³+b³=(a+b)(a²-ab+b²)

Here a=(2x)² ; b=(3y)²

⇒[(2x)²-(3y)²][((2x)²)²+(2x)².(3y)²+((3y)²)²]

⇒[4x²-9y²][16x⁴+4x².9y²+((9y²)²]

⇒[4x²-9y²][16x⁴+36x²y²+81y⁴]

⇒[(2x)²-(3y)²][16x⁴+36x²y²+81y⁴]

⇒(2x+3y)(2x-3y)(16x⁴+36x²y²+81y⁴)

(ii) $x^2-\frac{8}{x}$

⇒$\left(x^2-\frac{8}{x}\right)\times \frac{x}{x}$

⇒$x\left(x^2-\frac{8}{x}\right)\frac{1}{x}$

⇒$\left(x^3-x.\frac{8}{x}\right)\frac{1}{x}$

⇒$(x^3-8)\frac{1}{x}$

⇒$\left(x^3-2^3\right).\frac{1}{x}$

It is in the form of a³+b³=(a+b)(a²-ab+b²)

a=x ; b=2

⇒$\frac{1}{x}.(x+2)(x^2+2x+4)$

Question 12

(i) 250(a-b)³+2

⇒$(250(a-b)^3+2)\frac{2}{2}$

⇒$2\frac{(250(a-b)^3+2)}{2}$

⇒2(125(a-b)³+1)

⇒2(5³(a-b)³+1)

⇒2((5(a-b))³+1³)

It is in the form of a³+b³=(a+b)(a²-ab+b²)

Here a=5(a-b) ; b=1

⇒2[5(a-b)+1](5(a-b))²-5(a-b).1+1²]

⇒2[(5a-5b+1)][25(a²+b²-2ab)-5a+5b+1]

⇒2(5a-5b+1)(25a²+25b²-50ab-5a+5b+1)

(ii) 32a²x³-8b²x³-4a²y³+b²y³

⇒x³(32a²-8b²)-y³(4a²-b²)

⇒8x³(4a²-b²)-y³(4a²-b²)

⇒(4a²-b²)(8x³-y³)

⇒((2a)²-(b)²)((2x)³-(y)³)

⇒(2a+b)(2a-b)[(2x-y)(2x)²+2xy+y]

⇒(2a+b)(2a-b)(2x-y)(4x²+2xy+y²)

Question 13

(i) x²+y²

⇒(x³)³+(y³)³

It is in the form of a³+b³=(a+b)(a²-ab+b²)

Here a=x³ ; b=y³

∴(x³+y³)[(x³)³-x³y³+(y³)²]

⇒(x³+y³)(x⁶-x³y³+(y³)²)

It is in the form of a³+b³=(a+b)(a²-ab+b²)

⇒Here a=x ; b=y

∴⇒(x+y)(x²-xy+y²)(x⁶-x³y³+y⁶)

(ii) x⁶-7x³-8

⇒(x²)³-7x³-x³-8

⇒(x²)³-8x³+x³-2³

⇒[(x²)³-(2x)³]+(x³-2²)

⇒(x²-2x)[(x²)²+x².2x+(2x)²]+(x-2)(x²+2x+2²)

⇒(x²-2x²)(x⁴+2x³+4x²)+(x-2)(x²+2x+4)

⇒x(x-2)x²(x²+2x+4)+(x-2)(x²+2x+4)

Taking common factor as (x-2) and x²+2x+4

∴(x-2)(x²+2x+4)(x+1)(x²-x+1)