Factorize : a³- b³- c³ – 3abc

Factorize : a³ – b³ – c³– 3abc

Let’s term this equation : a³ – b³ – c³– 3abc ……………… (1)

By using the identity of :  a³ + b³ +  c³– 3abc, the above equation (1) can be factorized. 

Where

a= a

b= -b and 

c= -c 

For to solve the above-given equation : a³ – b³ – c³ – 3abc

The basic equation which is to be used shall be known which is :

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

Some of the facts and forms in which this equation can be written are be known regarding this equation such as :

• a³ + b³ +  c³ = 3abc ( where a + b + c = 0 )

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

• a³ + b³ +  c³ = (a+ b+ c )² – 3ab (a + b ) – 3bc (b +c ) – 3ca ( c +a ) – 6abc 

Identity  

a³ + b³ +  c³ – 3abc = (a+ b+ c ) ( a² + b² + c² – ab – bc – ca ) ……….(2 )

When considering the equation (2) in the form of equation (1) it becomes :

a³ + (- b³) + (- c³) – 3a (-b)(-c) = (a + ( – b ) + (- c ) ) ( a² + (-b²) + (- c² ) – a (-b) – (-b) (-c) – (-c) a

Then this equation becomes

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

Hence

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

For example  

 a = 3

b= 3

c= 3 

 Then : a³ – b³ – c³– 3abc  

= ( 3 – 3 – 3 ) ( 3² + 3² + 3² + 9 – 9 + 9 )

= (3) (27 +9 )

= (3) (36) 

= 108