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

Factorize each of the following expressions:

Question 13. 8x²y³-x⁵

Solution:

⇒ x²((2y)³ – x³)

⇒ x²(2y-x)((2y)²+2xy+x²)                           [x³-y³=(x-y)(x²+xy+y²)]

⇒ x²(2y-x)(4y²+2xy+x²)

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

Question 14. 1029 – 3x³

Solution:

⇒ 3 (343 – x³)

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

⇒ 3 (7-x)(7²+7x+x²)

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

Question 15. x⁶+y⁶

Solution:

⇒ (x²)³ + (y²)³

⇒ (x²+y²)((x²)²-(xy)²+(y²)²)

⇒ (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:

⇒ (xy)³ + 1³

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

⇒ (xy+1) (x²y² – xy +1)

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

Question 17. x⁴y⁴ – xy

Solution:

⇒ xy(x³y³ – 1)

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

⇒ xy (xy -1)((xy)² +xy +1²)

⇒ xy (xy-1)(x²y² + xy+1)       

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

Question 18. a¹²+b¹²

Solution:

⇒ (a⁴)³+(b⁴)³

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

⇒ (a⁴+b⁴)(a⁸-a⁴b⁴+b⁸)

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

Question 19. x³+6x²+12x+16

Solution:

⇒ x³+6x²+12x+8+8

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

⇒ (x+2)³+2³

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

⇒ (x+4)(x²+2²+4x-2x-4+4)

⇒(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:

⇒ (a³+b³)+1(a+b)

⇒ (a+b)(a²-ab+b²)+1(a+b)                            [a³+b³ = (a+b)(a²-ab+b²)]

⇒ (a+b)(a² -ab +b² +1)

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

Question 21. 

Solution:

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

Therefore, 

Question 22. 

Solution:

⇒ (a+b)³-8                             [a³+3a²b+3ab²+b³ = (a+b)³] 

⇒ (a+b)³ – 2³

⇒ (a+b-2) ((a+b)²+2(a+b)+2²)

⇒(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:

⇒ (2a)³ – b³ -2x(2a -b)

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

 ⇒ (2a-b)(4a² +2ab + b²-2x)

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

Question 24.

(i) 

Solution:

⇒ 

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

⇒ 173+127 

⇒ 300

(ii) 

Solution:

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

⇒ 

⇒ 1.2 – 0.2

⇒ 1.0

(iii)

Solution:

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

⇒ 

⇒ 155-55

⇒ 100