RD Sharma Class 9 Solutions Chapter 5 Exercise 5.4 (Updated for 2024)

RD Sharma Class 9 Solutions Chapter 5 Exercise 5.4

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 + 8b+ 64c− 24abc

Solution:

a3 + 8b+ 64c− 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)(a+4b+16c−2ab−8bc−4ac)

Therefore, a3 + 8b+ 64c− 24abc = (a+2b+4c)(a+4b+16c−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)+ (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 + 4y+ 9 z+ 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 + c−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 + c−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 + 25z+ xy/3 + 5yz – 5zx/3)

Question 5: 8x3 + 27y− 216z+ 108xyz

Solution:

8x3 + 27y− 216z+ 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) {4x+9y+36z−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+4x+9y−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 + c−3abc = (a + b + c)(a2+b2+c2−ab−bc−ca)

⇒ a3 + b3 + c−3abc = 0

or a3 + b3 + c=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 + c−3abc = (a + b + c)(a2+b2+c2−ab−bc−ca)

⇒ a3 + b3 + c−3abc = 0

(2x−3y)3 + (4z−2x) 3 + (3y−4z) 3 = 3(2x−3y)(4z−2x)(3y−4z)

This is the complete blog on RD Sharma Class 9 Solutions Chapter 5 Exercise 5.4. To know more about the CBSE Class 9 Maths exam, ask in the comments.

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