Factorize using Cubes of Binomial Identity

FACTORIZE USING CUBES OF BINOMIAL IDENTITY:

1. Cube of a Binomial:

 (a + b)³  = a³ + 3a²b + 3ab² + b³.

 ( a – b )³ = a³ – 3a²b + 3ab² – b^3.

2. Sum of Cubes:

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

              = (a + b)³ – 3ab( a + b )

3. Difference of Cubes:

a³ – b³ = ( a – b )(a² + ab + b² ).

            = ( a – b )³ + 3ab ( a – b )

 

Factorize the following:

 1)  (2x – 2/x)³

Answer:

 The given expression: (2x – 2/x)³ ; is in the form of : ( a – b )³  .

Here,  a = 2x ; b = 2/x .

Substituting a , b  values in the Cube of Binomial Identity :

 ( a – b )³ = a³ – 3a²b + 3ab² – b^3.

= (2x – 2/x)³

=  (2x)³ – 3 * (2x)² * (2/x) + 3 * (2x) * (2/x)² – (2/x)³

=  8x³ – 3 * 4x^{2 *} 2/x +  3* 2x * 4/x² – 8 / x³

=  8x³ – 24x + 24/x – 8/x³ .

 

2) 2x³ – 54

Answer:

In  the given expression: 2x³ – 54; if we take out number ‘2’ as common , the expression changes in to : 2 ( x³ – 27 ) = 2 ( x³ – 3³  ) as we know 27 = 3³  and the new expression is in the form of : Difference of Cubes.

 The Difference of Cubes Identity :  a³ – b³ = ( a – b )(a² + ab + b² ).

 Here, a = x ; b = 3 .

 a²  = x² ; b² = 3² = 9 ; ab = 3x .

Therefore, 2 ( x³ – 3³  )

= 2 ( x – 3 ) ( x² + 3x + 3² ).

= 2 (x-3)( x² + 3x + 9 ).

    ( OR )

Substituting a,b values in  :( a³ –  b³ ) =  ( a – b )³ + 3ab ( a – b )

Therefore , 2 ( x³ – 3³  ) becomes

= 2 (( x – 3 )³ + 3 * 3x ( x- 3 ) ).

3) If  x =1  + √ 2; then find ( x – 1/x )³ ?

Answer:

Here,  we need to find ( x – 1/x )³  ,  which is in the form of : ( a – b )³ .

First we have to solve the given expression :

 ( x – 1/x )³  = (  ( x-1) / x)³

Substituting the given ‘ x’value in the expression:

= ( ( 1 + √ 2 - 1)  / (1  + √ 2) )³

= ( √ 2 /(1  + √ 2)³ ) ; the denominator is in the form of  Cube of Binomial Identity: 

( a + b )³ = a³ + 3a²b + 3ab² + b³.

Here,(1  + √ 2)³ expression: a = 1 ; b= √ 2

= 1³ + 3* 1²* √ 2 + 3* 1 * (√ 2)² + (√ 2)³

= 1 + 3√ 2 + 6 + 2 √ 2 

= (5 √ 2 + 7 )  ; substituting in ( √ 2 /(1  + √ 2)³ ) , then

 ( √ 2 / (1  +√ 2 )³ ) = ( √ 2 /( 5√ 2 + 7); multiplying numerator and denominator with 

(5√ 2 -7 ),i.e: rationalizing. Hence :

= ( √ 2 * (5 √ 2 - 7) / (5 √ 2 + 7) *( 5 √ 2 -7) ); denominator is in the form of Difference of squares Identity : (a+b)(a-b) = a² – b² .

= ( (10 - 7√ 2 ) / ( 50 – 49)

=  ( (10 - 7√ 2) / 1 )

=  10 - 7√ 2.