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

### Factorize each of the following:

### Question 1. a^{3} + 8b^{3} + 64c^{3 }â€“ 24abc

**Solution: **

We know that

a

^{3}+ b^{3}+ c^{3}â€“ 3abc = (a + b + c) (a^{2}+ b^{2}+ c^{2}â€“ ab â€“ bc â€“ ca)= a

^{3}+ 8b^{3}+ 64c^{3}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) (a^{2}+ 4b^{2}+ 16c^{2}â€“ 2ab â€“ 8bc â€“ 4ca)

### Question 2. x^{3} â€“ 8y^{3} + 27z^{3} + 18xyz

**Solution: **

We can simplify the given equation as :

x

^{3}â€“ 8y^{3}+ 27z^{3}+ 18xyz= (x)

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

(x â€“ y + 3z) (x^{2}+ 4y^{2}+ 9z^{2}+ 2xy + 6yz â€“ 3zx)

### Question 3. 27x^{3} â€“ y^{3}â€“ z^{3} â€“ 9xyz

**Solution:**

27x

^{3}– y^{3}– z^{3}– 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) (9x^{2}+ y^{2}+ z^{2}+ 3xy â€“ yz + 3zx)

### Question **4. (1/27)x**^{3 }– y^{3 }+ 125z^{3 }+ 5xyz

^{3 }– y

^{3 }+ 125z

^{3 }+ 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)x^{2}+ y^{2}+25z^{2}+ (1/3xy) + 5yz – (5/3xz)]

### Question 5. 8x^{3} + 27y^{3} â€“ 216z^{3} + 108xyz

**Solution:**

8x

^{3}+ 27y^{3}â€“ 216z^{3}+ 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) (4x^{2}+ 9y^{2}+ 36z^{2}â€“ 6xy + 18yz + 12zx)

### Question 6. 125 + 8x^{3} â€“ 27y^{3} + 90xy

**Solution:**

125 + 8X

^{3}â€“ 27y^{3}+ 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 + 4x^{2}+ 9y^{2}â€“ 10x + 6xy + 15y)

### Question 7. 8x^{3} â€“ 125y^{3} + 180xy + 216

**Solution:**

8x

^{3}â€“ 125y^{3}+ 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) (4x^{2}+ 25y^{2}+ 36 + 10xy + 30y â€“ 12x)

### Question 8. Multiply:

**(i) x ^{2} +y^{2} + z^{2} â€“ xy + xz + yz by x + y â€“ z**

**(ii) x ^{2} + 4y^{2} + z^{2} + 2xy + xz â€“ 2yz by x- 2y-z**

**(iii) x ^{2} + 4y^{2} + 2xy â€“ 3x + 6y + 9 by x â€“ 2y + 3**

**(iv) 9x ^{2} + 25y^{2} + 15xy + 12x â€“ 20y + 16 by 3x â€“ 5y + 4**

**Solution:**

(i) (x^{2}+ y^{2}+ z^{2}â€“ xy + yz + zx) by (x + y â€“ z)= x

^{3}+y^{3}â€“ z^{3}+ 3xyz

(ii) (x^{2}+ 4y^{2}+ z^{2}+ 2xy + xz â€“ 2yz) by (x â€“ 2y â€“ z)= (x -2y-z) [x

^{2}+ (-2y)^{2}+ (-z)^{2}– (x * (- 2y)) â€“ ((-2y)* (z)) â€“ ((-z) (x))]= x

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

^{3}â€“ 8y^{3}â€“ z^{3}â€“ 6xyz

(iii) x^{2}+ 4y^{2}+ 2xy â€“ 3x + 6y + 9 by (x â€“ 2y + 3)= (x â€“ 2y + 3) (x

^{2}+ 4y^{2}+ 9 + 2xy + 6y â€“ 3x)= (x)

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

^{3}â€“ 8y^{3}+ 27 + 18xy

(iv) 9x^{2}+ 25y^{2}+ 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= 27x

^{3}â€“ 125y^{3}+ 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 x

^{3}+ y^{3}+ z^{3}= 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 x

^{3}+ y^{3}+ z^{3}= 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}

^{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 a

^{3}+ b^{3}+ c^{3}= 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 a^{3}+ b^{3}+ c^{3}= 3abc if a + b + c = 0}

### Question **13. 2**âˆš2a^{3} + 3âˆš3b^{3} + c^{3} – 3âˆš6abc

**Solution:**

= (âˆš2a)

^{3}+(âˆš3b)^{3}+c^{3}– 3 * âˆš2a * âˆš3b * c= (âˆš2a + âˆš3b +c)[(âˆš2a)

^{2}+(âˆš3b)^{2}+ c^{2}– (âˆš2a * âˆš3b) – (âˆš3b * c) – (c * âˆš2a)= (âˆš2a + âˆš3b +c)(2a

^{2}+ 3b^{2}+ c^{2}– âˆš6ab – âˆš3bc – âˆš2ca)

### Question **14. 3**âˆš3a^{3} – b^{3} – 5âˆš5c^{3} – 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)(3a^{2}+ b^{2}+ 5c^{2}+ âˆš3ab – âˆš5bc + âˆš15ca)

### Question **15. 2âˆš2 a**^{3} + 16âˆš2 b^{3} + c^{3} – 12abc

^{3}+ 16

**Solution:**

=(âˆš2a)

^{3}+ (2âˆš2b)^{3}+ c^{3}– (3 * âˆš2a * 2âˆš2b * c)=(âˆš2a + 2âˆš2b +c) [(âˆš2a)

^{2}+ (2âˆš2b)^{2}+ c^{2}– (âˆš2a* 2âˆš2b) – (2âˆš2b*c) – (c* âˆš2a)=(âˆš2a + 2âˆš2b +c)[2a

^{2}+ 8b^{2}+ c^{2}– 4ab – 2âˆš2bc – âˆš2ca]

### Question 16. Find the value of x^{3 }+ y^{3} â€“ 12xy + 64, when x + y = – 4

**Solution:**

x

^{3}+ y^{3}â€“ 12xy + 64x + y = -4

On Cubing both sides,

x

^{3}+ y^{3}+ 3 xy(x + y) = -64Substitute the value of (x + y)

â‡’ x

^{3}+ y^{3}+ 3xy * (-4) = -64â‡’ x

^{3}+ y^{3}â€“ 12xy + 64 = 0

## Please

Loginto comment...