SELINA Solution Class 9 Factorisation Chapter 5 Exercise 5D

Question 1

Factorise : a³ - 27

Sol:

a³ - 27 
= ( a )³ - ( 3 )³
= ( a - 3 )[ (a)² + a x 3 + (3)² ]        [ ∵ a³ - b³ = ( a - b )( a² + ab + b² )]
= ( a - 3 )[ a² + 3a + 9 ]

Question 2

Factorise : 1 - 8a³

Sol:

1 - 8a³
= (1)³ - (2a)³
= ( 1 - 2a )[ (1)² + 1 x 2a + (2a)² ]    
                            [ ∵ a³ - b³ = ( a - b )( a² + ab + b² )]
= ( 1 - 2a )[ 1 + 2a + 4a² ]

Question 3

Factorise : 64 - a³b³

Sol:

64 - a³b³ 
= (4)³ - (ab)³
= ( 4 - ab )[(4)² + 4 x ab + (ab)² ]            [ ∵ a³ - b³ = ( a - b )( a² + ab + b² )]
= ( 4 - ab )( 16 + 4ab + a²b² )

Question 4

Factorise : a⁶ + 27b³

Sol:

a⁶ + 27b³ 
= ( a² )³ + ( 3b )³
= ( a² + 3b )[ (a²)² - a² x 3b + (3b)² ]          [ ∵ a³ + b³ = ( a + b )( a² - ab + b² )]
= ( a² + 3b )[ a⁴ - 3a²b + 9b² ]

Question 5

Factorise : 3x⁷y - 81x⁴y⁴

Sol:

3x⁷y - 81x⁴y⁴ 

= 3xy( x⁶ - 27x³y³ )

= 3xy[ (x²)³ - ( 3xy )³ ]

= 3xy( x² - 3xy )[ (x²)² + x² x 3xy + (3xy)² ]          [ ∵ a³ - b³ = ( a -b )( a² + ab + b² )]

= 3xy( x² - 3xy )[ x⁴ + 3x³y + 9x²y² ]

= 3xy [ x( x + 3y) x²( x² + 3xy + 9y² ) ]

= 3x⁴y( x - 3y )( x² + 3xy + 9y² )

Question 6

Factorise : a³ - ${\displaystyle \frac{27}{a^3}}$

Sol:

a³ - ${\displaystyle \frac{27}{a^3}}$

= (a)³ - ${\displaystyle {\left(\frac{3}{a}\right)}^3}$

= ${\displaystyle (a-\frac{3}{a})[a^2+a\times \frac{3}{a}+{\left(\frac{3}{a}\right)}^2]}$          [ ∵ a³ +b³ = ( a -b )( a² + ab + b² )]

= ${\displaystyle (a-\frac{3}{a})(a^2+3+\frac{9}{a^2})}$

Question 7

Factorise : a³ + 0.064

Sol:

a³ + 0.064 
= (a)³ + (0.4)³
= ( a + 0.4 )[ (a)² - a x 0.4 + (0.4)² ]               [ ∵ a3 + b3 = ( a + b )( a² - ab + b² ) ]

= ( a + 0.4 )( a² - 0.4a + 0.16 )

Question 8

Factorise : a⁴ - 343a

Sol:

a⁴ - 343a 
= a( a³ - 7³ )
= a( a - 7 )[(a)² + a x 7 + (7)² ]                       [ ∵ a³ - b³ = ( a - b )( a² + ab + b² )]
= a( a - 7 )( a² + 7a + 49 )

Question 9

Factorise: (x - y)³ - 8x³

Sol:

( x - y )³ - 8x³
= ( x - y )³ - ( 2x )³
= ( x - y - 2x )[ (x - y)² + 2x(x - y) + (2x)² ]
  [ Using identity (a³ - b³) = (a - b)(a² + ab + b²) ]

= (- x - y ) [ x² + y² - 2xy + 2x² - 2xy + 4x² ]
=- ( x + y ) [ 7x² - 4xy + y² ]

 

Question 10

Factorise : ${\displaystyle \frac{{\left(8a\right)}^3}{27}-\frac{b^3}{8}}$

Sol:

${\displaystyle \frac{{\left(8a\right)}^3}{27}-\frac{b^3}{8}}$

= ${\displaystyle {\left(\frac{2a}{3}\right)}^3-{\left(\frac{b}{2}\right)}^3}$

= ${\displaystyle (\frac{2a}{3}-\frac{b}{2})[{\left(\frac{2a}{3}\right)}^2+\frac{2a}{3}\times \frac{b}{2}+{\left(\frac{b}{2}\right)}^2]}$

[ ∵ a³ - b³ = ( a - b )( a² + ab + b² )]

= ${\displaystyle (\frac{2a}{3}-\frac{b}{2})[\frac{{\left(4a\right)}^2}{9}+\frac{ab}{3}+\frac{b^2}{4}]}$

Question 11

Factorise : a⁶ - b⁶

Sol:

We know that,
a³ + b³ = ( a + b )( a² - ab + b² )                     ....(1)
a³ - b³ = ( a - b )( a² + ab + b² )                      ....(2)
a⁶ - b⁶
= ( a³)² - (b³)²
= ( a³ + b³ )( a³ - b³ )
= ( a + b )( a² - ab + b² )( a - b )( a² + ab + b² )     
[ From(1) and (2) ]
= ( a + b )( a - b )( a² - ab + b² )( a² + ab + b² )

Question 12

Factorise : a⁶ - 7a³ - 8

Sol:

We know that,
a³ + b³ = ( a + b )( a² - ab + b² )                     ....(1)
a³ - b³ = ( a - b )( a² + ab + b² )                      ....(2)
a⁶ - 7a³ - 8
= a⁶ - 8a³ + a³ - 8
= a³( a³ - 8) + 1( a³ - 8 )
= ( a³ + 1 )( a³ - 8 )
= ( a³ + 1³ )( a³ - 2³ )
= ( a + 1 )( a² - a + 1 )( a - 2 )( a² + 2a + 4 )       
[ From(1) and (2) ]
= ( a + 1 )( a - 2)( a² - a + 1 )( a² + 2a + 4 )

Question 13

Factorise : a³ - 27b³ + 2a²b - 6ab²

Sol:

a³ - 27b³ + 2a²b - 6ab²
We know that,
a³ - b³ = ( a - b )( a² + ab + b² )                 ....(1)
a³ - 27b³ + 2a²b - 6ab²
= (a)³ - (3b)³ + 2ab( a - 3b )
= ( a - 3b )[ a² + a x 3b + (3b)² ] + 2ab( a - 3b )        [From(1)]
= ( a - 3b )[ a² + 3ab + 9b² ] + 2ab( a - 3b )
= ( a - 3b )[ a² + 3ab + 9b² + 2ab ]
= ( a - 3b )[ a² + 5ab + 9b² ]

Question 14

Factorise : 8a³ - b³ - 4ax + 2bx

Sol:

We know that,
a³ - b³ = ( a - b )( a² + ab + b² )         .....(1)
8a³ - b³ - 4ax + 2bx
= [ (2a)³ - (b)³ ] - 2 x ( 2a - b )
= ( 2a - b )[ (2a)² + 2a x b + (b)² ] - 2 x ( 2a - b )    [ From(1) ]
= ( 2a - b )[ 4a² + 2ab + b² ] - 2 x ( 2a - b )
= ( 2a - b )[ 4a² + 2ab + b² - 2x ]

Question 15

Factorise : a - b - a³ + b³

Sol:

we know that,
a³ - b³ = ( a - b )( a² + ab + b² )         ....(1)
a - b - a³ + b³
= a - b - ( a³ - b³ )
= ( a - b ) - ( a - b )[ a² + ab + b² ]          [ From (1) ]
= ( a - b )[ 1 - a² - ab - b² ]

Question 16

Factorise :  2x³ + 54y³ - 4x - 12y

Sol:

2x³ + 54y³ - 4x - 12y

= 2 ( x³ + 27y³ - 2x - 6y )

= 2 [ { (x)³+ (3y)³} - 2(x  + 3y) ]
[ Using identity (a³ +  b³) = (a + b)(a² - ab + b²) ]
= 2[ {(x + 3y)(x² - 3xy + 9y²)} - 2(x + 3y) ]
= 2(x + 3y)(x² - 3xy + 9y² - 2)

Question 17

Factorise : 1029 - 3x³

Sol:

1029 - 3x³

= 3( 343 - x³ )

= 3( 7³ - x³ )

= 3( 7 - x )( 7² + 7x + x² )

= 3( 7 - x )( 49 + 7x + x² )

Question 18.1

Show that : 13³ - 5³ is divisible by 8

Sol:

 ( 13³ - 5³ )
[ Using identity (a³ - b³) = (a - b)(a² + ab + b²)]

= ( 13 - 5 )( 13² + 13 × 5 + 5² )
= 8( 169 + 65 + 25 )
Therefore, the number is divisible by 8.

Question 18.2

Show that : 35³ + 27³ is divisible by 62

Sol:

(35³ + 27³)
[Using identity (a³ + b³)=(a + b)(a² - ab + b²)]
= ( 35 + 27 )( 35² + 35× 27 + 27² )
= 62 × ( 35² + 35 × 27 + 27² )
Therefore, the number is divisible by 62.

Question 19

Evaluate : 
${\displaystyle \frac{5.67\times 5.67\times 5.67+4.33\times 4.33\times 4.33}{5.67\times 5.67-5.67\times 4.33+4.33\times 4.33}}$

Sol:

Let a = 5.67 and b = 4.33
Then,
${\displaystyle \frac{5.67\times 5.67\times 5.67+4.33\times 4.33\times 4.33}{5.67\times 5.67-5.67\times 4.33+4.33\times 4.33}}$

= ${\displaystyle \frac{a\times a\times a+b\times b\times b}{a\times a-a\times b+b\times b}}$

= ${\displaystyle \frac{a^3+b^3}{a^2-ab+b^2}}$

= ${\displaystyle \frac{(a+b)(a^2-ab+b^2)}{a^2-ab+b^2}}$

= a + b 

= 5.67 + 4.33 

= 10