# Class 9 RD Sharma Solutions – Chapter 5 Factorisation of Algebraic Expressions- Exercise 5.3

• Last Updated : 19 Jan, 2021

### Question 1. Factorize 64a3+125b3+240a2b+300ab2

Solution:

As we know that, a3+b3+3a2b+3ab2 = (a+b)3

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So, the above expression can be written as:

(4a)3+(5b)3+3(4a)2(5b)+3(4a)(5b)2

(4a+5b)3

(4a+5b)(4a+5b)(4a+5b)

Hence, 64a3+125b3+240a2b+300ab2 = (4a+5b)(4a+5b)(4a+5b)

### Question 2. Factorize 125x3-27y3-225x2y+135xy2

Solution:

As we know that, a3 – b3-3a2b+3ab2 = (a-b)3

So, the above expression can be written as:

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

(5x-3y)3

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

Hence, 125x3-27y3-225x2y+135xy2 = (5x-3y)(5x-3y)(5x-3y)

### Question 3. Factorize 8/27x3+1+4/3x2+2x

Solution:

As we know that, a3+b3+3a2b+3ab2 = (a+b)3

So, the above expression can be written as:

(2/3x)3+13+3(2/3x)2(1)+ 3(2/3x)(1)

(2/3x+1)3

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

Hence, 8/27x3+1+4/3x2+2x = (2/3x+1) (2/3x+1) (2/3x+1)

### Question 4. Factorize 8x3+27y3+36x2y+54xy2

Solution:

As we know that, a3+b3+3a2b+3ab2 = (a+b)3

So, the above expression can be written as:

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

(2x+3y)3

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

Hence, 8x3+27y3+36x2y+54xy2 = (2x+3y)(2x+3y)(2x+3y)

### Question 5. Factorize a3-3a2b+3ab2-b3+8

Solution:

As we know that, a3 – b3-3a2b+3ab2 = (a-b)3

So, the above expression can be written as:

(a-b)3 + (2)3

(a-b+2)[(a-b)2-(a-b)(2)+22]

Use a3+b3=(a+b)(a2-ab+b2

(a-b+2)(a2+b2-2ab-2a+2b+4)

Hence, a3-3a2b+3ab2-b3+8 = (a-b+2)(a2+b2-2ab-2a+2b+4)

### Question 6. Factorize x3+8y3+6x2y+12xy2

Solution:

As we know that, a3+b3+3a2b+3ab2 = (a+b)3

So, the above expression can be written as:

x3+(2y)3+3(x)2(2y)+3(x)(2y)2

(x+2y)3

(x+2y)(x+2y)(x+2y)

Hence, x3+8y3+6x2y+12xy2 =  (x+2y)(x+2y)(x+2y)

### Question 7. Factorize 8x3+y3+12x2y+6xy2

Solution:

As we know that, a3+b3+3a2b+3ab2 = (a+b)3

So, the above expression can be written as:

(2x)3 +y3+3(2x)2y+3(2x)(y)2

(2x+y)3

(2x+y)(2x+y)(2x+y)

Hence, 8x3+y3+12x2y+6xy2 = (2x+y)(2x+y)(2x+y)

### Question 8. Factorize 8a3+27b3+36a2b+54ab2

Solution:

As we know that, a3+b3+3a2b+3ab2 = (a+b)3

So, the above expression can be written as:

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

(2a+3b)3

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

Hence, 8a3+27b3+36a2b+54ab2 =  (2a+3b) (2a+3b) (2a+3b)

### Question 9. Factorize 8a3-27b3-36a2b+54ab2

Solution:

As we know that, a3-b3-3a2b+3ab2 = (a-b)3

So, the above expression can be written as:

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

(2a-3b)3

(2a-3b) (2a-3b) (2a-3b)

Hence, 8a3-27b3-36a2b+54ab2 = (2a-3b) (2a-3b) (2a-3b)

### Question 10. Factorize x3 – 12x(x-4)-64

Solution:

As we know that, a3-b3-3a2b+3ab2 = (a-b)3

So, the above expression can be written as:

x3 -12x2 +48x -43

x3 -43-3(x)2(4) +3(x)(4)2

(x-4)3

(x-4) (x-4) (x-4)

Hence,  x3 – 12x(x-4)-64 = (x-4) (x-4) (x-4)

### Question 11. Factorize a3x3 -3a2bx2 +3ab2x-b3

Solution:

As we know that, a3-b3-3a2b+3ab2 = (a-b)3

So, the above expression can be written as:

(ax)3 -3(ax)2(b)+3(ax)(b)-b3

(ax-b)3

(ax-b) (ax-b) (ax-b)

Hence, a3x3 -3a2bx2 +3ab2x-b3  = (ax-b) (ax-b) (ax-b)

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