## 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

- RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.1
- 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 5.3
- RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.4
- RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions VSAQS
- RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions MCQS

Factorize each of the following expressions:

Question 1.

p^{3} + 27

Solution:

We know that a^{3} + b^{3} = (a + b) (a^{2} – ab + b^{2})

a^{3} – b^{3} = (a – b) (a^{2} + aft + b^{2})

p^{3} + 21 = (p)^{3} + (3)^{3
}= (p + 3) (p^{2}– p x 3 + 3^{2})

= (p + 3) (p^{2} – 3p + 9)

Question 2.

y^{3} + 125

Solution:

y^{3} + 125 = (p)^{3} + (5)^{3
}= (p + 5) (p^{2} – 5y + 5^{2})

= (P + 5) (p^{2} – 5y + 25)

Question 3.

1 – 21a^{3
}Solution:

1 – 21a^{3} = (1)^{3} – (3a)^{3
}= (1 – 3a) [1^{2} + 1 x 3a + (3a)^{2}]

= (1 – 3a) (1 + 3a + 9a^{2})

Question 4.

8x^{3}y^{3} + 27a^{3 }

Solution:

8x^{3}y^{3} + 27a^{3
}= (2xy + 3a) [(2xy)^{2} – 2xy x 3a + (3a)^{2}]

= (2xy + 3a) (4x^{2}y – 6xya + 9a^{2})

Question 5.

64a^{3} – b^{3
}Solution:

64a^{3} – b^{3} = (4a)^{3} – (b)^{3
}= (4a – b) [(4a)^{2} + 4a x b + (b)^{2}]

= (4a – b) (16a^{2} + 4ab + b^{2})

Question 6.

Solution:

Question 7.

10x^{4}– 10xy^{4
}Solution:

I0x^{4}y- 10xy^{4} = 10xy(x^{3} -y^{3})

= 10xy(x – y) (x^{2} + xy + y^{2})

Question 8.

54x^{6}y + 2x^{3}y^{4
}Solution:

54 x^{6}y + 2x^{3}y^{4} = 2x^{3}y(27x^{3} + y^{3})

= 2x^{3}y[(3x)^{3} + (y)^{3}]

= 2x^{3}y(3x + y) [(3x)^{2} -3x x y + y^{2}]

= 2x^{3}y(3x + y) (9x^{2} -3xy + y^{2})

Question 9.

32a^{3} + 108b^{3 }

Solution:

32a^{3} + 108b^{3
}= 4(8a^{3} + 27b^{3}) = 4 [(2a)^{3} + (3 b)^{3}]

= 4(2a + 3b) [(2a)^{2} – 2a x 3b + (3b)^{2}]

= 4(2a + 3b) (4a^{2} – 6ab + 9b^{2})

Question 10.

(a – 2b)^{3} – 512b^{3 }

Solution:

(a – 2b)^{3} – 512b^{3
}= (a – 2b)^{3} – (8b)^{3
}= (a – 2b- 8b) [(a – 2b)^{2} + (a – 2b) x 8b + (8b)^{2}]

= (a – 10b) [a^{2} + 4b^{2} – 4ab + 8ab – 16b^{2} + 64b^{2}]

= (a – 10b) (a^{2} + 4ab + 52b^{2})

Question 11.

8x^{2}y^{3} – x^{5
}Solution:

8x^{2}y^{3} – x^{5} = x^{2}(8y^{3} – X^{3})

= x^{2}(2y)^{3} – (x)^{3}]

= x^{2}[(2y – x) (2y)^{2} + 2y x x + (x)^{2}]

= x^{2}(2y – x) (4y^{2} + 2xy + x^{2})

Question 12.

1029 -3x^{3
}Solution:

1029 – 3X^{3} = 3(343 – x^{3}) ‘

= 3 [(7)^{3} – (x)^{3}]

= 3(7 – x) (49x + 7x + x^{2})

Question 13.

x^{3}y^{3}+ 1

Solution:

x^{3}y^{3} + 1 = (xy)^{3} + (1)^{3
}= (xy + 1) [(xy)^{2} – xy x 1 + (1)^{2}]

= (xy + 1) (x^{2}y^{2} – xy + 1)

= (xy + 1) (x^{2}y – xy + 1)

Question 14.

x^{4}y^{4} – xy

Solution:

x^{4}y^{4} – xy = xy(x^{3}y^{3} – 1)

= xy[(xy^{3}-(1)^{3}]

= xy (xy – 1) [x^{2}y^{2} + 2xy + 1]

Question 15.

a^{3} + b^{3} + a + b

Solution:

a^{3} + b^{3} + a + b

= (a + b) (a^{2} – ab + b^{2}) + 1 (a + b)

= (a + b) (a^{2} – ab + b^{2} + 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) [a^{2}* + *b^{2}* + 2ab + 2a ^{2} – *2ab + 2ab – 2b

^{2}+ 4a

^{2}– 8ab + 4b

^{2}]

= (3b – a) [7a

^{2}– 6ab + 3b

^{2}]

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 [x^{2} + 4x + 4 – (x^{2} + 2x – 2x – 4) + x^{2 }4x + 4]

= 2x[x^{2} + 4x + 4- x^{2}-2x + 2x + 4+ x^{2}– 4x + 4]

= 2x[x^{2} + 12]

Question 19.

x^{6} +y^{6
}Solution:

x^{6} + y^{6 }= (x^{2})^{3} + (y^{2})^{3
}= (x^{2} + y^{2}) [x^{4} – x^{2}y^{2} + y^{4}]

Question 20.

a^{12} + b^{12}

Solution:

a^{12} + b^{12} = (a^{4})^{3} + (b^{4})^{3
}= (a^{4} + b^{4}) [(a^{4})^{2} – a^{4}b^{4} + (b^{4})^{2}]

= (a^{4} + b^{4}) (a^{8} – a^{4}b^{4} + b^{8})

Question 21.

x^{3} + 6x^{2} + 12x + 16

Solution:

x^{3} + 6x^{2} + 12x + 16

= (x)^{3} + 3.x^{2}.2 + 3.x.4 + (2)^{3} + 8 {∵ a^{3} + 3a^{2}b + 3ab^{2} +b^{3} = (a + b)^{3}}

= (x + 2)^{3} + 8 = (x + 2)^{3} + (2)^{3
}= (x + 2 + 2) [(x + 2)^{2} – (x + 2) x 2 + (2)^{2}] {∵ a^{3} + b^{2} = (a + b) (a^{2} – ab + b^{2}}

= (x + 4) (x^{2} + 4x + 4 – 2x – 4 + 4)

= (x + 4) (x^{2} + 2x + 4)

Question 22.

Solution:

Question 23.

a^{3} + 3a^{2}b + 3ab^{2} + b^{3} – 8

Solution:

a^{3} + 3a^{2}b + 3ab^{2} + b^{3} – 8

= (a + b)^{3} – (2)^{3
}= (a + b -2)[(a + b)^{2} + (a +b)x2 + (2)^{2}]

= (a + b-2) (a^{2} + b^{2} + 2ab + 2a + 2b + 4)

= (a + b – 2) (a^{2} + b^{2} + 2ab + 2(a + b) + 4]

= (a + b – 2) [(a + b)^{2} + 2(a + b) + 4}

Question 24.

*8a ^{3} *– b

^{3}– 4ax + 2bx

Solution:

8a

^{3}– b

^{3}– 4ax + 2bx

(2a)

^{3}– (b)

^{3}– 2x(2a – b)

= (2a-b)[(2a)

^{2}+ 2a x b + (b)

^{2}]- 2x(2a-b)

= (2a – b) [4a

^{2}+ 2ab + b

^{2}] – 2x(2a – b)

= (2a – b) [4a

^{2}+ 2ab + b

^{2}– 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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