RD Sharma Solutions -Ex-5.2, Factorization Of Algebraic Expressions, Class 9, Maths

# RD Sharma Solutions -Ex-5.2, Factorization Of Algebraic Expressions, Class 9, Maths - Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

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Q 1 . p3+27

SOLUTION :

= p3+33                                       ∵[a3+b3=(a+b)(a2−ab+b2)]

= (p + 3)(p² – 3p – 9)

∴ p3+27 =(p + 3)(p² – 3p – 9)

Q 2 . y3+125

SOLUTION :

=  y3+5                                      ∵[a3+b3=(a+b)(a2−ab+b2)]

= (y+5)(y2−5y+52)

= (y+5)(y2−5y+25)

∴ y3+125 = (y+5)(y2−5y+25)

Q 3 . 1−27a3

SOLUTION :

= (1)3−(3a)3

= (1−3a)(12+1×3a+(3a)2)                  ∵[a3−b3=(a−b)(a2+ab+b2)]

= (1−3a)(12+3a+9a2)

∴ 1−27a= (1−3a)(12+3a+9a2)

Q 4 . 8x3y3+27a3

SOLUTION :

= (2xy)3+(3a)3

= (2xy+3a)((2xy)2−2xy×3a+(3a)2)                        ∵[a3+b3=(a+b)(a2−ab+b2)]

= (2xy+3a)(4x2y2−6xya+9a2)

∴ 8x3y3+27a3 = (2xy+3a)(4x2y2−6xya+9a2)

Q 5 . 64a3−b3

SOLUTION :

= (4a)3−b3

= (4a−b)((4a)2+4a×b+b2)                                 ∵[a3−b3=(a−b)(a2+ab+b2)]

=(4a−b)(16a2+4ab+b2)

∴  64a3−b3 =(4a−b)(16a2+4ab+b2)

Q 6 .

SOLUTION :

∵ [x3−y3=(x−y)(x2+xy+y2)]

Q 7 . 10x4y−10xy4

SOLUTION  :

= 10xy(x3−y3)

= 10xy(x−y)(x2+xy+y2)                          ∵[x3−y3=(x−y)(x2+xy+y2)]

∴ 10x4y−10xy4  = 10xy(x−y)(x2+xy+y2)

Q 8 . 54x6y+2x3y4

SOLUTION  :

= 2x3y(27x3+y3)

= 2x3y((3x)3+y3)

= 2x3y(3x+y)((3x)2−3x×y+y2)                  ∵[a3+b3=(a+b)(a2−ab+b2)]

=2x3y(3x+y)(9x2−3xy+y2)

∴ 54x6y+2x3y4 =2x3y(3x+y)(9x2−3xy+y2)

Q 9 . 32a3+108b3

SOLUTION  :

= 4(8a3+27b3)

= 4((2a)3+(3b)3)

= 4[(2a+3b)((2a)2−2a×3b+(3b)2)]                                                ∵[a3+b3=(a+b)(a2−ab+b2)]

=4(2a+3b)(4a2−6ab+9b2)

∴ 32a3+108b3 =4(2a+3b)(4a2−6ab+9b2)

Q 10 . (a−2b)3−512b3

SOLUTION  :

= (a−2b)3−(8b)3

= (a−2b−8b)((a−2b)2+(a−2b)8b+(8b)2)                                    ∵[a3−b3=(a−b)(a2+ab+b2)]

=(a−10b)(a2+4b2−4ab+8ab−16b2+64b2)

=(a−10b)(a2+52b2+4ab)

∴  (a−2b)3−512b3 =(a−10b)(a2+52b2+4ab)

Q 11 . (a+b)3−8(a−b)3

SOLUTION  :

= (a+b)3−[2(a−b)]3

= (a+b)3−[2a−2b]3

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

=(a+b−2a+2b)(a2+b2+2ab+(a+b)(2a−2b)+(2a−2b)2)

=(a+b−2a+2b)(a2+b2+2ab+2a2−2ab+2ab−2b2+(2a−2b)2)

=(3b−a)(3a2+2ab−b2+(2a−2b)2)

=(3b−a)(3a2+2ab−b2+4a2+4b2−8ab)

=(3b−a)(3a2+4a2−b2+4b2−8ab+2ab)

=(3b−a)(7a2+3b2−6ab)

∴  (a+b)3−8(a−b)3 =(3b−a)(7a2+3b2−6ab)

Q 12 . (x+2)3+(x−2)3

SOLUTION  :

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

=2x(x2+4x+4−(x+2)(x−2)+x2−4x+4)

=2x(2x2+8−(x2−22))                           [∵(a+b)(a−b)=a2−b2]

=2x(2x2+8−x2+4)

=2x(x2+12)

∴ (x+2)3+(x−2)3 =2x(x2+12)

Q 13 .  8x2y3−x5

SOLUTION  :

= x2((2y)3−x3)

= x2(2y−x)((2y)2+2y×x+x2)     [∵x3−y3=(x−y)(x2+xy+y2)]

= x2(2y−x)(4y2+2xy+x2)

∴ 8x2y3−x= x2(2y−x)(4y2+2xy+x2)

Q  14 . 1029 – 3x3

SOLUTION  :

= 3(343−x3)

= 3((7)3−x3)

= 3(7−x)(72+7x+x2)                                    [∵a3−b3=(a−b)(a2+ab+b2)]

=3(7−x)(49+7x+x2)

∴ 1029 – 3x3 =3(7−x)(49+7x+x2)

Q 15 . x6+y6

SOLUTION  :

= (x2)3+(y2)3

= (x2+y2)((x2)2−x2y2+(y2)2)

= (x2+y2)(x4−x2y2+y4)                                                [∵a3+b3=(a+b)(a2−ab+b2)]

∴ x6+y6 = (x2+y2)(x4−x2y2+y4)

Q 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)

∴ x3y3+1 = (xy+1)(x2y2−xy+1)

Q 17 . x4y4−xy

SOLUTION  :

= xy(x3y3−1)

= xy((xy)3−13)

= xy(xy−1)((xy)2+xy×1+12)                                             ∵[x3−y3=(x−y)(x2+xy+y2)]

=xy(xy−1)(x2y2+xy+1)

∴ x4y4−xy = xy(xy−1)(x2y2+xy+1)

Q 18 . a12+b12

SOLUTION  :

= (a4)3+(b4)3

= (a4+b4)((a4)2−a4×b4+(b4)2)                                                ∵[a3+b3=(a+b)(a2−ab+b2)]

=(a4+b4)(a8−a4b4+b8)

∴ a12+b12 = (a4+b4)(a8−a4b4+b8)

Q 19 . x3+6x2+12x+16

SOLUTION  :

= x3+6x2+12x+8+8

= x3+3×x2×2+3×x×22+23+8

= (x+2)3+8                                              [∵a3+3a2b+3ab2+b3=(a+b)3]

= (x+2)3+23

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

=(x+2+2)(x2+4+4x−2x−4+4)                       [∵(a+b)2=a2+b2+2ab]

=(x+4)(x2+4+2x)

∴ x3+6x2+12x+16 = (x+4)(x2+4+2x)

Q 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)

∴ a3+b3+a+b = (a+b)(a2−ab+b2+1)

Q 21 .

SOLUTION  :

[∵a3−b3=(a−b)(a2+ab+b2)]

Q 22 . a3+3a2b+3ab2+b3−8

SOLUTION  :

= (a+b)3−8                                                                [∵a3+3a2b+3ab2+b3=(a+b)3]

=(a+b)3−23

=(a+b−2)((a+b)2+(a+b)×2+22)

= (a + b – 2)(a² + 2ab + b² + 2a + 2b + 4)

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

Q 23 . 8a3−b3−4ax+2bx

SOLUTION  :

= (2a)3−b3−2x(2a−b)

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

=(2a−b)(4a2+2ab+b2−2x)

∴ 8a3−b3−4ax+2bx= (2a−b)(4a2+2ab+b2−2x)

Q 24 . i .

SOLUTION  :

∵[a3+b3=(a+b)(a2−ab+b2)]

=(173+127)

=300

Q 24 . ii .

SOLUTION  :

[∵a3−b3=(a−b)(a2+ab+b2)]

=(1.2−0.2)

=1.0

Q 24 . iii

SOLUTION  :

[∵a3−b3=(a−b)(a2+ab+b2)]

= (155 – 55)

= 100

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