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# RD Chapter 5- Factorisation of Algebraic Expressions Ex-5.3 Interview Questions Answers

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

64a3 + 125b3 + 240a2b+ 300ab2
= (4a)3 + (5b)3 + 3 x (4a)2 x5b + 3(4a) + (5b)2
= (4a + 5b)3
= (4a + 5b) (4a + 5b) (4a + 5b)

Question 2 : 125x3 – 27y3 – 225x2y + 135xy2

125x3 – 27y3 – 225x2y+ 135xy2
= (5x)3 – (3y)3 – 3 x (5x)2 x(3y) + 3- x 5x x (3y)2
= (5x – 3y)3
(5x – 3y) (5x – 3y) (5x – 3y)

Question 3 :

Question 4 : 8x3 + 27y3 + 36x2y + 54xy2

8x3 + 27y3 + 16x2y +54xy2
= (2x)3 + (3y)3 + 3 x (2x)2 x3y  +  3 x 2x x (3y)2
= (2x + 3y)3
= (2x + 3y) (2x + 3y) (2x + 3y)

Question 5 : a3 – 3a2b + 3ab2 – b3 + 8

a3 – 3a2b + 3ab2 – b3 +8
= (a – b)3 + (2)3
= (a – b + 2) [(a -b)2– (a – b) x 2 + (2)2]
= (a- b + 2) (a2 + b2 -2ab – 2a + 2b + 4)

Question 6 : x3 + 8y3 + 6x2y + 12xy2

x3 + 8y3 + 6x2y +12xy2
= (x)3 + (2y)3 + 3 x x2x 2y + 3 x xx (2y)2
= (x + 2y)3
= (x + 2y) (x + 2y) (x + 2y)

Question 7 : 8x3 + y3 + 12x2y+ 6xy2

8x3 + y3 + 12x2y +6xy2
= (2x)3 + (y)3 + 3 x (2x)2 xy + 3 x 2x x y2
= (2x + y)3
= (2x + y) (2x + y) (2x + y)

Question 8 : 8a3 + 27b3 + 36a2b+ 54ab2

8a3 + 27b3 + 16a2b +54ab2
= (2a)3 + (3b)3 + 3 x (2a)x3b + 3 x 2a x (3b)2
= (2a + 3b)3
= (2a + 3b) (2a + 3b) (2a + 3b)

Question 9 : 8a3 – 27b3 – 36a2b+ 54ab2

8a3 – 27b3 – 36a2b +54ab2
= (2a)3 – (3b)3 – 3 x (2a)2 x3b + 3 x 2a x (3b)2
= (2a – 3b))3
= (2a – 3b) (2a – 3b) (2a – 3b)

Question 10 : x3 – 12x(x – 4) – 64

x3 – 12x(x – 4) – 64
= x3 – 12x2 + 48x – 64
= (x)3 – 3 x x2 x 4 + 3 x x x (4)2– (4)3
= (x – 4)3
= (x – 4) (x – 4) (x – 4)

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