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Class 9 RD Sharma Solutions – Chapter 5 Factorisation of Algebraic Expressions- Exercise 5.4

  • Last Updated : 28 Mar, 2021
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Factorize each of the following:

Question 1. a3 + 8b3 + 64c3 – 24abc

Solution: 

We know that

a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)

= a3 + 8b3 + 64c3 – 24abc

= (a)3 + (2b)3 + (4c)3 – (3 × a × 2b × 4c)



= (a + 2b + 4c) [(a)2 + (2b)2 + (4c)2 -(a × 2b)– (2b × 4c)– (4c × a)]2

= (a + 2b + 4c) (a2 + 4b2 + 16c2 – 2ab – 8bc – 4ca) 

Question 2. x3 – 8y3 + 27z3 + 18xyz

Solution: 

We can simplify the given equation as :

x3 – 8y3 + 27z3 + 18xyz

= (x)3 + (-2y)3 + (3z)3 – ( 3 * x * (-2y) * (3 z))

= (x – y + 3z) (x2 + 4y2 + 9z2 + 2xy + 6yz – 3zx) 

Question 3. 27x3 – y3– z3 – 9xyz         

Solution:



27x3 – y3 – z3 – 9xyz

= (3x)3 + (-y)3 + (-z)3 – (3 * 3x * (-y) (-z))

= (3x – y – z) [(3x)2 + (-y)2 + (-z)2 – (3x * (-y)) – ((-y) (-z))- (- z × 3x)]

= (3x – y – z) (9x2 + y2 + z2 + 3xy – yz + 3zx)

Question 4. (1/27)x3 – y3 + 125z3 + 5xyz

Solution:

=(1/3x)3 + -(y)3 +(5z)3 – 3*(1/3x) * (-y) * (5z)

=((1/3x) – y +5z)[(1/3x)2 + (-y)2 + (5z)2 -((1/3x)*-y) – (-y * 5z) – (5z * (1/3x))

=((1/3x) – y +5z)[(1/9)x2 + y2 +25z2 + (1/3xy) + 5yz – (5/3xz)]

Question 5. 8x3 + 27y3 – 216z3 + 108xyz

Solution:

8x3 + 27y3 – 216z3 + 108xyz



= (2x)3 + (3y)3 + (6z)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) [(5)2 + (2x)2 + (-3y)2 – (5 * 2x) – (2x * (-3y)) – ((-3y) * 5)]

= (5 + 2x – 3y) (25 + 4x2 + 9y2– 10x + 6xy + 15y)

Question 7. 8x3 – 125y3 + 180xy + 216

Solution:

8x3 – 125y3 + 180xy + 216

= (2x)3 + (-5y)3 + (6)3 – 3 * 2x *(-5y) * 6

= (2x – 5y + 6) [(2x)2 + (-5y)2 + (6)2 – 2x *(-5y) – (-5y) * 6 – 6 * 2x]

= (2x – 5y + 6) (4x2 + 25y2 + 36 + 10xy + 30y – 12x)

Question 8. Multiply:

(i) x2 +y2 + z2 – xy + xz + yz by x + y – z

(ii) x2 + 4y2 + z2 + 2xy + xz – 2yz by x- 2y-z

(iii) x2 + 4y2 + 2xy – 3x + 6y + 9 by x – 2y + 3

(iv) 9x2 + 25y2 + 15xy + 12x – 20y + 16 by 3x – 5y + 4

Solution:

(i) (x2 + y2 + z2 – xy + yz + zx) by (x + y – z)

= x3 +y3 – z3 + 3xyz



(ii) (x2 + 4y2 + z2 + 2xy + xz – 2yz) by (x – 2y – z)

= (x -2y-z) [x2 + (-2y)2 + (-z)2 – (x * (- 2y)) – ((-2y)* (z)) – ((-z) (x))]

= x3 + (-2y)3 + (-z)3 – 3x * (-2y) * (-z)

= x3 – 8y3 – z3 – 6xyz

(iii) x2 + 4y2 + 2xy – 3x + 6y + 9 by  (x – 2y + 3)

= (x – 2y + 3) (x2 + 4y2 + 9 + 2xy + 6y – 3x)

= (x)3 + (-2y)3 + (3)3 – (3 * x * (-2y) x 3)

= x3 – 8y3 + 27 + 18xy

(iv) 9x2 + 25y2 + 15xy + 12x – 20y + 16 by (3x – 5y + 4)

= (3x -5y + 4) [(3x)2 + (-5y)2 + (4)2 – 3x * (-5y) +(-5y x 4) + (4 × 3x)]

= (3x)3 + (-5y)3 + (4)3 – 3 * 3x *(-5y) * 4

= 27x3 – 125y3 + 64 + 180xy

Question 9. (3x – 2y)3 + (2y – 4z)3 + (4z – 3x)3

Solution:

(3x – 2y)3 + (2y – 4z)3 + (4z – 3x)3

∵ 3x – 2y + 2y – 4z + 4z – 3x = 0

∴ (3x – 2y)3 + (2y – 4z)3 + (4z – 3x)3

= 3(3x – 2y) (2y – 4z) (4z – 3x)               {∵ x3 + y3 + z3 = 3xyz if x + y + z = 0}

Question 10. (2x – 3y)3 + (4z – 2x)3 + (3y – 4z)3

Solution:

(2x – 3y)3 + (4z – 2x)3 + (3y – 4z)3

∵ 2x – 3y + 4z – 2x + 3y – 4z = 0



∴ (2x – 3y)3 + (4z – 2x)3 + (3y – 4z)3

= 3(2x – 3y) (4z – 2x) (3y – 4z)                {∵ x3 + y3 + z3 = 3xyz if x + y + z = 0}

Question 11. [(x/2)+y +(z/3)]3 + [(x/3) -(2y/3) +z]3 + [(-5x/6)-(y/3)-(4z/3)]3

Solution:

[(x/2)+y +(z/3)]3 + [(x/3) -(2y/3) +z]3 + [(-5x/6)-(y/3)-(4z/3)]3

∵ (x/2) + y +(z/3) +(x/3) -(2y/3) + z – (5x/6) -(y/3) – (4z/3) =0

∴ [(x/2)+y +(z/3)]3 + [(x/3) -(2y/3) +z]3 + [(-5x/6)-(y/3)-(4z/3)]3

= 3[(x/2)+y +(z/3)] [(x/3) -(2y/3) +z] [(-5x/6)-(y/3)-(4z/3)]     {∵ a3 + b3 + c3 = 3abc if a + b + c = 0}

Question 12. (a – 3b)3 + (3b – c)3 + (c – a)3

Solution:

(a- 3b)3 + (3b – c)3 + (c – a)3

∵ a – 3b + 3b – c + c – a = 0

∴ (a – 3b)3 + (3b – c)3 + (c – a)3

= 3(a – 3b) (3b – c) (c – a)                       {∵ a3 + b3 + c3 = 3abc if a + b + c = 0}

Question 13. 2√2a3 + 3√3b3 + c3 – 3√6abc

Solution:

= (√2a)3 +(√3b)3 +c3 – 3 * √2a * √3b * c

= (√2a + √3b +c)[(√2a)2 +(√3b)2 + c2 – (√2a * √3b) – (√3b * c) – (c * √2a)

= (√2a + √3b +c)(2a2 + 3b2 + c2 – √6ab – √3bc – √2ca)

Question 14. 3√3a3 – b3 – 5√5c3 – 3√15abc

Solution: 

= (√3a)3 + (-b)3 +(-√5c)3 – 3*√3a* (-b) *(-√5c)

= (√3a – b – √5c) [(√3a)2 + (-b)2 +(-√5c)2 – (√3a* -b) – (-b * (-√5c)) – (-√5c* √3a)

= (√3a – b – √5c)(3a2 + b2 + 5c2 + √3ab – √5bc + √15ca)



Question 15. 2√2 a3  + 16√2 b3 + c3 – 12abc

Solution:

=(√2a)3 + (2√2b)3 + c3 – (3 * √2a * 2√2b * c)

=(√2a + 2√2b +c) [(√2a)2 + (2√2b)2 + c2 – (√2a* 2√2b) – (2√2b*c) – (c* √2a)

=(√2a + 2√2b +c)[2a2 + 8b2 + c2 – 4ab – 2√2bc – √2ca]

Question 16. Find the value of x3 + y3 – 12xy + 64, when x + y = – 4

Solution:

x3 + y3 – 12xy + 64

x + y = -4

On Cubing both sides,

x3 + y3 + 3 xy(x + y) = -64

Substitute the value of (x + y)

⇒ x3 + y3 + 3xy * (-4) = -64

⇒ x3 + y3 – 12xy + 64 = 0

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