Factorizing using Cubes of Trinomial Identity

 CUBES OF TRINOMIAL IDENTITY :

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

 If  ( a + b + c ) = 0 ; then a³ + b³ + c³ = 3 abc .

 Factorize the expressio using Algebraic Identities:

1)  ( a - b)³ + ( b – c )³ + ( c – a )³

Answer:

The given expression: ( a - b)³ + ( b – c )³ + ( c – a )³ ; is in the form of  : a³ + b³ + c³

Here;  a = a – b ; b = b – c ; c = c – a .

By using the Cubes of Trinomial Identity :

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

 If a + b + c = 0 ,then  a³ + b³ + c³ = 3 abc

substituting  a, b , c values in the above Identity we get:

a + b + c = a – b +  b – c + c – a  = 0; 

( a - b)³ + ( b – c )³ + ( c – a )³

=  0 + 3 ( a-b)*( b-c) * (c-a)

= 3 (a-b)(b-c)(c-a). 

2)  x³ + y³ - 3xy +1.

Answer:

The given expression : x³ + y³ - 3xy +1 , can be re-write as;

= x³ + y³ + 1³ – 3* x * y * 1 ; which is in the form of

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

Here, a =x ; b = y ; c = 1 ; a²  = x²  ; b² = y² ; c² = 1 ; ab= xy ; bc = y ; ca = x

Therefore ;  x³ + y³ + 1 – 3 xy

= ( x + y + 1 ) (x² + y² + 1 – xy – y – x )

= ( x + y + 1 ) (x² + y² – xy –y – x + 1).

 

3) 8x³ + 27y³ – 90xy + 125

Answer:

The given expression : 8x³ + 27y³ – 90xy + 125 , can be rewrite as :

=   8x³ + 27y³ + 125 – 90xy

=  ( 2x)³  + ( 3y)³ + 5³  - 3 * 2x * 3y * 5 ; which is in the form of

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

Here, a = 2x ; b = 3y ; c= 5 ; a² = 4x² ; b² = 9y² ; c² = 25

ab = 6xy ; bc = 15y ; ca = 10x.

Therefore ;  8x³ + 27y³ + 125 – 90xy

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

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