# RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.2

## RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.2

Other Exercises

Factorize each of the following expressions:
Question 1.
p3 + 27
Solution:
We know that a3 + b3 = (a + b) (a2 – ab + b2)
a3 – b3 = (a – b) (a2 + aft + b2)
p3 + 21 = (p)3 + (3)3
= (p + 3) (p2– p x 3 + 32)
= (p + 3) (p2 – 3p + 9)

Question 2.
y3 + 125
Solution:
y3 + 125 = (p)3 + (5)3
= (p + 5) (p2 – 5y + 52)
= (P + 5) (p2 – 5y + 25)

Question 3.
1 – 21a3
Solution:
1 – 21a3 = (1)3 – (3a)3
= (1 – 3a) [12 + 1 x 3a + (3a)2]
= (1 – 3a) (1 + 3a + 9a2)

Question 4.
8x3y3 + 27a3
Solution:
8x3y3 + 27a3
= (2xy + 3a) [(2xy)2 – 2xy x 3a + (3a)2]
= (2xy + 3a) (4x2y – 6xya + 9a2)

Question 5.
64a3 – b3
Solution:
64a3 – b3 = (4a)3 – (b)3
= (4a – b) [(4a)2 + 4a x b + (b)2]
= (4a – b) (16a2 + 4ab + b2)

Question 6. Solution: Question 7.
10x4– 10xy4
Solution:
I0x4y- 10xy4 = 10xy(x3 -y3)
= 10xy(x – y) (x2 + xy + y2)

Question 8.
54x6y + 2x3y4
Solution:
54 x6y + 2x3y4 = 2x3y(27x3 + y3)
= 2x3y[(3x)3 + (y)3]
= 2x3y(3x + y) [(3x)2 -3x x y + y2]
= 2x3y(3x + y) (9x2 -3xy + y2)

Question 9.
32a3 + 108b3
Solution:
32a3 + 108b3
= 4(8a3 + 27b3) = 4 [(2a)3 + (3 b)3]
= 4(2a + 3b) [(2a)2 – 2a x 3b + (3b)2]
= 4(2a + 3b) (4a2 – 6ab + 9b2)

Question 10.
(a – 2b)3 – 512b3
Solution:
(a – 2b)3 – 512b3
= (a – 2b)3 – (8b)3
= (a – 2b- 8b) [(a – 2b)2 + (a – 2b) x 8b + (8b)2]
= (a – 10b) [a2 + 4b2 – 4ab + 8ab – 16b2 + 64b2]
= (a – 10b) (a2 + 4ab + 52b2)

Question 11.
8x2y3 – x5
Solution:
8x2y3 – x5 = x2(8y3 – X3)
= x2(2y)3 – (x)3]
= x2[(2y – x) (2y)2 + 2y x x + (x)2]
= x2(2y – x) (4y2 + 2xy + x2)

Question 12.
1029 -3x3
Solution:
1029 – 3X3 = 3(343 – x3)                       ‘
= 3 [(7)3 – (x)3]
= 3(7 – x) (49x + 7x + x2)

Question 13.
x3y3+ 1
Solution:
x3y3 + 1 = (xy)3 + (1)3
= (xy + 1) [(xy)2 – xy x 1 + (1)2]
= (xy + 1) (x2y2 – xy + 1)
= (xy + 1) (x2y – xy + 1)

Question 14.
x4y4 – xy
Solution:
x4y4 – xy = xy(x3y3 – 1)
= xy[(xy3-(1)3]
= xy (xy – 1) [x2y2 + 2xy + 1]

Question 15.
a3 + b3 + a + b
Solution:
a3 + b3 + a + b
= (a + b) (a2 – ab + b2) + 1 (a + b)
= (a + b) (a2 – ab + b2 + 1)

Question 16.
Simplify: Solution:  Question 17.
(a + b)3 – 8(a – b)3
Solution:
(a + b)3 – 8(a – b)3
= (a + b)2 – (2a – 2b)3
= (a+ b – 2a + 2b) [(a + b)2 + (a + b) (2a-2b) + (2a – 2b)2)]
= (3b – a) [a2 + b2 + 2ab + 2a22ab + 2ab – 2b2 + 4a2 – 8ab + 4b2]
= (3b – a) [7a2 – 6ab + 3b2]

Question 18.
(x + 2)3 + (x- 2)3
Solution:
(x + 2)3 + (x – 2)3
= (x + 2 + x – 2) [(x + 2)2 – (x + 2) (x – 2) + (x – 2)2]
= 2x [x2 + 4x + 4 – (x2 + 2x – 2x – 4) + x4x + 4]
= 2x[x2 + 4x + 4- x2-2x + 2x + 4+ x2– 4x + 4]
= 2x[x2 + 12]

Question 19.
x6 +y6
Solution:
x6 + y= (x2)3 + (y2)3
= (x2 + y2) [x4 – x2y2 + y4]

Question 20.
a12 + b12
Solution:
a12 + b12 = (a4)3 + (b4)3
= (a4 + b4) [(a4)2 – a4b4 + (b4)2]
= (a4 + b4) (a8 – a4b4 + b8)

Question 21.
x3 + 6x2 + 12x + 16
Solution:
x3 + 6x2 + 12x + 16
= (x)3 + 3.x2.2 + 3.x.4 + (2)3 + 8           {∵ a3 + 3a2b + 3ab2 +b3 = (a + b)3}
= (x + 2)3 + 8 = (x + 2)3 + (2)3
= (x + 2 + 2) [(x + 2)2 – (x + 2) x 2 + (2)2] {∵ a3 + b2 = (a + b) (a2 – ab + b2}
= (x + 4) (x2 + 4x + 4 – 2x – 4 + 4)
= (x + 4) (x2 + 2x + 4)

Question 22. Solution: Question 23.
a3 + 3a2b + 3ab2 + b3 – 8
Solution:
a3 + 3a2b + 3ab2 + b3 – 8
= (a + b)3 – (2)3
= (a + b -2)[(a + b)2 + (a +b)x2 + (2)2]
= (a + b-2) (a2 + b2 + 2ab + 2a + 2b + 4)
= (a + b – 2) (a2 + b2 + 2ab + 2(a + b) + 4]
= (a + b – 2) [(a + b)2 + 2(a + b) + 4}

Question 24.
8a3 – b3 – 4ax + 2bx
Solution:
8a3 – b3 – 4ax + 2bx
(2a)3 – (b)3 – 2x(2a – b)
= (2a-b)[(2a)2 + 2a x b + (b)2]- 2x(2a-b)
= (2a – b) [4a2 + 2ab + b2] – 2x(2a – b)
= (2a – b) [4a2 + 2ab + b2 – 2x]

Hope given RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.2 are helpful to complete your math homework.

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