# Class 9 RD Sharma Solutions – Chapter 5 Factorisation of Algebraic Expressions- Exercise 5.2 | Set 2

Last Updated : 11 Feb, 2021

### Question 13. 8x2y3-x5

Solution:

â‡’ x2((2y)3 – x3)

â‡’ x2(2y-x)((2y)2+2xy+x2)                           [x3-y3=(x-y)(x2+xy+y2)]

â‡’ x2(2y-x)(4y2+2xy+x2)

Therefore, 8x3y3-x5 = x2(2y-x)(4y2+2xy+x2)

### Question 14. 1029 – 3x3

Solution:

â‡’ 3 (343 – x3)

â‡’ 3(73 – x3)                           [x3-y3=(x-y)(x2+xy+y2)]

â‡’ 3 (7-x)(72+7x+x2)

Therefore, 1029 – 3x3 = 3 (7-x)(72+7x+x2)

### Question 15. x6+y6

Solution:

â‡’ (x2)3 + (y2)3

â‡’ (x2+y2)((x2)2-(xy)2+(y2)2)

â‡’ (x2+y2)(x4-x2y2+y4)                           [x3+y3=(x+y)(x2-xy+y2)]

Therefore,  x6+y6 = (x2+y2)(x4-x2y2+y4)

### Question 16. x3y3+1

Solution:

â‡’ (xy)3 + 13

â‡’(xy+1)((xy)2-xy+12)                            [x3+y3=(x+y)(x2-xy+y2)]

â‡’ (xy+1) (x2y2 – xy +1)

Therefore, x3y3+1 = (xy+1) (x2y2 – xy +1)

### Question 17. x4y4 – xy

Solution:

â‡’ xy(x3y3 – 1)

â‡’ xy ((xy)3 – 13)                           [x3-y3=(x-y)(x2+xy+y2)]

â‡’ xy (xy -1)((xy)2 +xy +12)

â‡’ xy (xy-1)(x2y2 + xy+1)

Therefore, x4y4 – xy = xy (xy-1)(x2y2 + xy+1)

### Question 18. a12+b12

Solution:

â‡’ (a4)3+(b4)3

â‡’ (a4+b4)((a4)2-a4b4+(b4)2)                           [x3+y3=(x+y)(x2-xy+y2)]

â‡’ (a4+b4)(a8-a4b4+b8)

Therefore, a12+b12 = (a4+b4)(a8-a4b4+b8)

### Question 19. x3+6x2+12x+16

Solution:

â‡’ x3+6x2+12x+8+8

â‡’ x3+3*x2*2+3*x*22+23+8                            [a3+3a2b+3ab2+b3 = (a+b)3]

â‡’ (x+2)3+23

â‡’ (x+2+2)((x+2)2-2(x+2)+22)                         [x3+y3=(x+y)(x2-xy+y2)]

â‡’ (x+4)(x2+22+4x-2x-4+4)

â‡’(x+4)(x2+4+2x)                                            [(a+b)2=a2+b2+2ab]

Therefore, x3+6x2+12x+16 = (x+4)(x2+4+2x)

### Question 20. a3+b3+a+b

Solution:

â‡’ (a3+b3)+1(a+b)

â‡’ (a+b)(a2-ab+b2)+1(a+b)                            [a3+b3 = (a+b)(a2-ab+b2)]

â‡’ (a+b)(a2 -ab +b2 +1)

Therefore, a3+b3+a+b = (a+b)(a2 -ab +b2 +1)

Solution:

â‡’

â‡’

â‡’

â‡’

â‡’

Therefore,

### Question 22.

Solution:

â‡’ (a+b)3-8                             [a3+3a2b+3ab2+b3 = (a+b)3

â‡’ (a+b)3 – 23

â‡’ (a+b-2) ((a+b)2+2(a+b)+22)

â‡’(a+b-2)(a2+2ab+b2+2a+2b+4)

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

### Question 23. 8a3– b3 -4ax+2bx

Solution:

â‡’ (2a)3 – b3 -2x(2a -b)

â‡’ (2a-b)((2a)2 +2ab + b2) -2x(2a-b)                              [x3-y3=(x-y)(x2+xy+y2)]

â‡’ (2a-b)(4a2 +2ab + b2-2x)

Therefore, 8a3-b3-4ax+2bx = (2a-b)(4a2 +2ab + b2-2x)

### (i)

Solution:

â‡’

â‡’ \frac{(173+127)(173^2-173*127+127^2)}{173^2-173*127+127^2}                   [a3+b3=(a+b)(a2+b2-ab)]

â‡’ 173+127

â‡’ 300

### (ii)

Solution:

â‡’ \frac{1.2^3-0.2^3}{1.2^2+1.2*0.2+0.2^2}                                                            [a3-b3=(a-b)(a2+b2+ab)]

â‡’

â‡’ 1.2 – 0.2

â‡’ 1.0

### (iii)

Solution:

â‡’ \frac{153^3-55^3}{155^2+155*55+55^2}                                                               [a3-b3=(a-b)(a2+b2+ab)]

â‡’

â‡’ 155-55

â‡’ 100

Previous
Next