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

• Last Updated : 28 Mar, 2021

### Question 1. a3 + 8b3 + 64c3 â€“ 24abc

Solution:

We know that

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

= a3 + 8b3 + 64c3 â€“ 24abc

= (a)3 + (2b)3 + (4c)3 â€“ (3 Ã— a Ã— 2b Ã— 4c)

= (a + 2b + 4c) [(a)2 + (2b)2 + (4c)2 -(a Ã— 2b)â€“ (2b Ã— 4c)â€“ (4c Ã— a)]2

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

### Question 2. x3 â€“ 8y3 + 27z3 + 18xyz

Solution:

We can simplify the given equation as :

x3 â€“ 8y3 + 27z3 + 18xyz

= (x)3 + (-2y)3 + (3z)3 â€“ ( 3 * x * (-2y) * (3 z))

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

### Question 3. 27x3 â€“ y3â€“ z3 â€“ 9xyz

Solution:

27x3 – y3 – z3 – 9xyz

= (3x)3 + (-y)3 + (-z)3 â€“ (3 * 3x * (-y) (-z))

= (3x â€“ y â€“ z) [(3x)2 + (-y)2 + (-z)2 â€“ (3x * (-y)) â€“ ((-y) (-z))- (- z Ã— 3x)]

= (3x – y â€“ z) (9x2 + y2 + z2 + 3xy â€“ 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 â€“ 216z3 + 108xyz

Solution:

8x3 + 27y3 â€“ 216z3 + 108xyz

= (2x)3 + (3y)3 + (6z)3 â€“ 3 Ã— (2x) (3y) (-6z)

= (2x + 3y â€“ 6z) [(2x)2 + (3y)2 + (-6z)2 â€“ (2x * 3y) â€“ (3y * (-6z)) â€“ ((-6z) * 2x)]

= (2x + 3y â€“ 6z) (4x2 + 9y2 + 36z2 â€“ 6xy + 18yz + 12zx)

### Question 6. 125 + 8x3 â€“ 27y3 + 90xy

Solution:

125 + 8X3 â€“ 27y3 + 90xy

= (5)3 + (2x)3 + (-3y)3 â€“ [3 * 5 * 2x * (-3y)]

= (5 + 2x â€“ 3y) [(5)2 + (2x)2 + (-3y)2 â€“ (5 * 2x) â€“ (2x * (-3y)) â€“ ((-3y) * 5)]

= (5 + 2x â€“ 3y) (25 + 4x2 + 9y2â€“ 10x + 6xy + 15y)

### Question 7. 8x3 â€“ 125y3 + 180xy + 216

Solution:

8x3 â€“ 125y3 + 180xy + 216

= (2x)3 + (-5y)3 + (6)3 â€“ 3 * 2x *(-5y) * 6

= (2x â€“ 5y + 6) [(2x)2 + (-5y)2 + (6)2 â€“ 2x *(-5y) â€“ (-5y) * 6 â€“ 6 * 2x]

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

### Question 8. Multiply:

(i) x2 +y2 + z2 â€“ xy + xz + yz by x + y â€“ z

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

(iii) x2 + 4y2 + 2xy â€“ 3x + 6y + 9 by x â€“ 2y + 3

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

Solution:

(i) (x2 + y2 + z2 â€“ xy + yz + zx) by (x + y â€“ z)

= x3 +y3 â€“ z3 + 3xyz

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

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

= x3 + (-2y)3 + (-z)3 â€“ 3x * (-2y) * (-z)

= x3 â€“ 8y3 â€“ z3 â€“ 6xyz

(iii) x2 + 4y2 + 2xy â€“ 3x + 6y + 9 by  (x â€“ 2y + 3)

= (x â€“ 2y + 3) (x2 + 4y2 + 9 + 2xy + 6y â€“ 3x)

= (x)3 + (-2y)3 + (3)3 â€“ (3 * x * (-2y) x 3)

= x3 â€“ 8y3 + 27 + 18xy

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

= (3x -5y + 4) [(3x)2 + (-5y)2 + (4)2 â€“ 3x * (-5y) +(-5y x 4) + (4 Ã— 3x)]

= (3x)3 + (-5y)3 + (4)3 â€“ 3 * 3x *(-5y) * 4

= 27x3 â€“ 125y3 + 64 + 180xy

### Question 9. (3x â€“ 2y)3 + (2y â€“ 4z)3 + (4z â€“ 3x)3

Solution:

(3x â€“ 2y)3 + (2y â€“ 4z)3 + (4z â€“ 3x)3

âˆµ 3x â€“ 2y + 2y â€“ 4z + 4z â€“ 3x = 0

âˆ´ (3x â€“ 2y)3 + (2y â€“ 4z)3 + (4z â€“ 3x)3

= 3(3x â€“ 2y) (2y â€“ 4z) (4z â€“ 3x)               {âˆµ x3 + y3 + z3 = 3xyz if x + y + z = 0}

### Question 10. (2x â€“ 3y)3 + (4z â€“ 2x)3 + (3y â€“ 4z)3

Solution:

(2x â€“ 3y)3 + (4z â€“ 2x)3 + (3y â€“ 4z)3

âˆµ 2x â€“ 3y + 4z â€“ 2x + 3y â€“ 4z = 0

âˆ´ (2x â€“ 3y)3 + (4z â€“ 2x)3 + (3y â€“ 4z)3

= 3(2x â€“ 3y) (4z â€“ 2x) (3y â€“ 4z)                {âˆµ 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

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

âˆ´ [(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)]     {âˆµ a3 + b3 + c3 = 3abc if a + b + c = 0}

### Question 12. (a â€“ 3b)3 + (3b â€“ c)3 + (c â€“ a)3

Solution:

(a- 3b)3 + (3b â€“ c)3 + (c â€“ a)3

âˆµ a â€“ 3b + 3b â€“ c + c â€“ a = 0

âˆ´ (a â€“ 3b)3 + (3b â€“ c)3 + (c â€“ a)3

= 3(a â€“ 3b) (3b â€“ c) (c â€“ a)                       {âˆµ a3 + b3 + c3 = 3abc if a + b + c = 0}

### Question 13. 2âˆš2a3 + 3âˆš3b3 + c3 – 3âˆš6abc

Solution:

= (âˆš2a)3 +(âˆš3b)3 +c3 – 3 * âˆš2a * âˆš3b * c

= (âˆš2a + âˆš3b +c)[(âˆš2a)2 +(âˆš3b)2 + c2 – (âˆš2a * âˆš3b) – (âˆš3b * c) – (c * âˆš2a)

= (âˆš2a + âˆš3b +c)(2a2 + 3b2 + c2 – âˆš6ab – âˆš3bc – âˆš2ca)

### Question 14. 3âˆš3a3 – b3 – 5âˆš5c3 – 3âˆš15abc

Solution:

= (âˆš3a)3 + (-b)3 +(-âˆš5c)3 – 3*âˆš3a* (-b) *(-âˆš5c)

= (âˆš3a – b – âˆš5c) [(âˆš3a)2 + (-b)2 +(-âˆš5c)2 – (âˆš3a* -b) – (-b * (-âˆš5c)) – (-âˆš5c* âˆš3a)

= (âˆš3a – b – âˆš5c)(3a2 + b2 + 5c2 + âˆš3ab – âˆš5bc + âˆš15ca)

### Question 15. 2âˆš2 a3  + 16âˆš2 b3 + c3 – 12abc

Solution:

=(âˆš2a)3 + (2âˆš2b)3 + c3 – (3 * âˆš2a * 2âˆš2b * c)

=(âˆš2a + 2âˆš2b +c) [(âˆš2a)2 + (2âˆš2b)2 + c2 – (âˆš2a* 2âˆš2b) – (2âˆš2b*c) – (c* âˆš2a)

=(âˆš2a + 2âˆš2b +c)[2a2 + 8b2 + c2 – 4ab – 2âˆš2bc – âˆš2ca]

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

Solution:

x3 + y3 â€“ 12xy + 64

x + y = -4

On Cubing both sides,

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

Substitute the value of (x + y)

â‡’ x3 + y3 + 3xy * (-4) = -64

â‡’ x3 + y3 â€“ 12xy + 64 = 0

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