Ex-5.2, Factorization Of Algebraic Expressions, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics PDF Download

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+5³                                       ∵[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)

∴ 1−27a³ = (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−x⁵ = 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