RD Sharma Solutions -Ex-5.3, Factorization Of Algebraic Expressions, Class 9, Maths

# RD Sharma Solutions -Ex-5.3, Factorization Of Algebraic Expressions, Class 9, Maths - Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

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Q 1 . 64a3+125b3+240a2b+300ab2

SOLUTION  :

= (4a)3+(5b)3+3(4a)2(5b)+3(4a)(5b)                                      [∵a3+b3+3a2b+3ab2=(a+b)3]

= (4a+5b)3

= (4a+5b)(4a+5b)(4a+5b)

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

Q 2 . 125x3−27y3−225x2y+135xy2

SOLUTION  :

= (5x)3−(3y)3−3(5x)2(3y)+3(5x)(3y)2                       [∵a3−b3−3a2b+3ab2=(a−b)3]

=(5x−3y)3

=(5x−3y)(5x−3y)(5x−3y)

∴ 125x3−27y3−225x2y+135xy= (5x−3y)(5x−3y)(5x−3y)

Q 3 .

SOLUTION  :

[∵x3+b3+3x2b+3xb2=(x+b)3]

Q 4 . 8x3+27y3+36x2y+54xy2

SOLUTION  :

= (2x)3+(3y)3+3×(2x)2×3y+3×(2x)(3y)2

= (2x+3y)3                                               [ ∵ a³ + b³ + 3a²b + 3ab² = (a + b) ³]

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

∴ 8x3+27y3+36x2y+54xy= (2x+3y)(2x+3y)(2x+3y)

Q 5 . a3−3a2b+3ab2−b3+8

SOLUTION  :

= (a−b)3+23                                        [∵a3−b3−3a2b+3ab2=(a−b)3]

=(a−b+2)((a−b)2−(a−b)2+22)                               ∵[a3+b3=(a+b)(a2−ab+b2)]

=(a−b+2)(a2+b2−2ab−2(a−b)+4)

=(a−b+2)(a2+b2−2ab−2a+2b+4)

∴ a3−3a2b+3ab2−b3+8 = (a−b+2)(a2+b2−2ab−2a+2b+4)

Q 6 . x3+8y3+6x2y+12xy2

SOLUTION  :

= (x)3+(2y)3+3×x2×2y+3×x×(2y)2

= (x+2y)3                                  [∵x3+y3+3x2y+3xy2=(x+y)3]

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

∴ x3+8y3+6x2y+12xy= (x+2y)(x+2y)(x+2y)

Q 7 . 8x3+y3+12x2y+6xy2

SOLUTION  :

= (2x)3+(y)3+3×(2x)2×y+3(2x)×y2

= (2x+y)3                                 [∵a3+b3+3a2b+3ab2=(a+b)3]

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

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

Q 8 . 8a3+27b3+36a2b+54ab2

SOLUTION  :

= (2a)3+(3b)3+3×(2a)2×3b+3×2a×(3b)2

= (2a+3b)3                                             [∵a3+b3+3a2b+3ab2=(a+b)3]

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

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

Q 9 . 8a3−27b3−36a2b+54ab2

SOLUTION  :

= (2a)3−(3b)3−3×(2a)2×3b+3×2a×(3b)2

= (2a–3b)3                                              [∵a3−b3−3a2b+3ab2=(a−b)3]

=(2a–3b)(2a–3b)(2a–3b)

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

Q 10 . x3−12x(x−4)−64

SOLUTION  :

= x3−12x2+48x−64

= x3−3×x2×4+3×42×x−43

= (x−4)3                                     [∵a3−b3−3a2b+3ab2=(a−b)3]

=(x−4)(x−4)(x−4)

∴ x3−12x(x−4)−64 =(x−4)(x−4)(x−4)

Q 11 . a3x3−3a2bx2+3ab2x−b3

SOLUTION  :

= (ax)3−3(ax)2×b+3(ax)×b2−b3

= (ax−b)3                                   [∵a3−b3−3a2b+3ab2=(a−b)3]

=(ax−b)(ax−b)(ax−b)

∴ a3x3−3a2bx2+3ab2x−b3 =(ax−b)(ax−b)(ax−b)

The document RD Sharma Solutions -Ex-5.3, Factorization Of Algebraic Expressions, Class 9, Maths - Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9 is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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## RD Sharma Solutions for Class 9 Mathematics

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## RD Sharma Solutions for Class 9 Mathematics

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