RD Sharma Class 9 Solutions Chapter 5 Exercise 5.4:The formula to success required constant practice along with the best book. Here we present you the RD Sharma Solutions Class 9 Maths. Not only it is easy to understand, but all the answers are credible as they are designed by subject matter experts. RD Sharma Class 9 Solutions Chapter 5 Exercise 5.4 can be your go-to maths guide and help you ace your Maths exam.
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RD Sharma Class 9 Solutions Chapter 5 Exercise 5.4
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Factorize each of the following expressions:
Question 1: a3 + 8b3 + 64c3 − 24abc
Solution:
a3 + 8b3 + 64c3 − 24abc
= (a)3 + (2b) 3 + (4c) 3− 3×a×2b×4c
[Using a3+b3+c3−3abc = (a+b+c)(a2+b2+c2−ab−bc−ca)]
= (a+2b+4c)(a2+(2b)2 + (4c)2−a×2b−2b×4c−4c×a)
= (a+2b+4c)(a2 +4b2 +16c2 −2ab−8bc−4ac)
Therefore, a3 + 8b3 + 64c3 − 24abc = (a+2b+4c)(a2 +4b2 +16c2 −2ab−8bc−4ac)
Question 2: x 3 − 8y 3+ 27z3 + 18xyz
Solution:
= x3 − (2y) 3 + (3z) 3 − 3×x×(−2y)(3z)
= (x + (−2y) + 3z) (x2 + (−2y)2 + (3z) 2 −x(−2y)−(−2y)(3z)−3z(x))
[using a3+b3+c3−3abc = (a+b+c)(a2+b2+c2−ab−bc−ca)]
=(x −2y + 3z)(x2 + 4y2 + 9 z2 + 2xy + 6yz − 3zx)
Question 3: 27x 3 − y 3– z3 – 9xyz
Solution:
27x 3 − y 3– z3 – 9xyz
= (3x) 3 − y 3– z3 – 3(3xyz)
[Using a3 + b3 + c3 −3abc = (a + b + c)(a2+b2+c2−ab−bc−ca)]
Here a = 3x, b = -y and c = -z
= (3x – y – z){ (3x)2 + (- y)2 + (– z)2 + 3xy – yz + 3xz)}
= (3x – y – z){ 9x2 + y2 + z2 + 3xy – yz + 3xz)}
Question 4: 1/27 x3 − y3 + 125z3 + 5xyz
Solution:
1/27 x3 − y3 + 125z3 + 5xyz
= (x/3)3+(−y)3 +(5z)3 – 3 x/3 (−y)(5z)
[Using a3 + b3 + c3 −3abc = (a + b + c)(a2+b2+c2−ab−bc−ca)]
= (x/3 + (−y) + 5z)((x/3)2 + (−y)2 + (5z) 2 –x/3(−y) − (−y)5z−5z(x/3))
= (x/3 −y + 5z) (x^2/9 + y2 + 25z2 + xy/3 + 5yz – 5zx/3)
Question 5: 8x3 + 27y3 − 216z3 + 108xyz
Solution:
8x3 + 27y3 − 216z3 + 108xyz
= (2x) 3 + (3y) 3 +(−6y) 3 −3(2x)(3y)(−6z)
= (2x+3y+(−6z)){ (2x)2+(3y) 2+(−6z) 2 −2x×3y−3y(−6z)−(−6z)2x}
= (2x+3y−6z) {4x2 +9y2 +36z2 −6xy + 18yz + 12zx}
Question 6: 125 + 8x3 − 27y3 + 90xy
Solution:
125 + 8x3 − 27y3 + 90xy
= (5)3 + (2x) 3 +(−3y) 3 −3×5×2x×(−3y)
= (5+2x+(−3y)) (52 +(2x) 2 +(−3y) 2 −5(2x)−2x(−3y)−(−3y)5)
= (5+2x−3y)(25+4x2 +9y2 −10x+6xy+15y)
Question 7: (3x−2y)3 + (2y−4z) 3 + (4z−3x) 3
Solution:
(3x−2y)3 + (2y−4z) 3 + (4z−3x) 3
Let (3x−2y) = a, (2y−4z) = b , (4z−3x) = c
a + b + c= 3x−2y+2y−4z+4z−3x = 0
We know, a3 + b3 + c3 −3abc = (a + b + c)(a2+b2+c2−ab−bc−ca)
⇒ a3 + b3 + c3 −3abc = 0
or a3 + b3 + c3 =3abc
⇒ (3x−2y)3 + (2y−4z) 3 + (4z−3x) 3 = 3(3x−2y)(2y−4z)(4z−3x)
Question 8: (2x−3y)3 + (4z−2x) 3 + (3y−4z) 3
Solution:
(2x−3y)3 + (4z−2x) 3 + (3y−4z) 3
Let 2x – 3y = a , 4z – 2x = b , 3y – 4z = c
a + b + c= 2x−3y+4z−2x+3y−4z = 0
We know, a3 + b3 + c3 −3abc = (a + b + c)(a2+b2+c2−ab−bc−ca)
⇒ a3 + b3 + c3 −3abc = 0
(2x−3y)3 + (4z−2x) 3 + (3y−4z) 3 = 3(2x−3y)(4z−2x)(3y−4z)
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