Question

what will be the derivation of this formula?
a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)​

Answer 1

Answer:

Prove a³ + b³ + c³ - 3abc = (a+b+c) (a ² + b² + c² - ab - bc - ac).

Good question,

Here is your perfect answer!

Solve LHS,

= (a³+b³) + c³ - 3abc

= {(a+b)³ - 3ab(a+b)} + c³ - 3abc

Let a+b be n.

= (n³ - 3abn + c³ - 3abc)

= (n³ + c³) - 3abc - 3abn

= (n+c) (n² - nc + c²) - 3ab(n+c)

= (n+c) (n² - nc + c² - 3ab)

Put n = a+b,

= (a+b+c) (a ² + b² + 2ab - (a+b) c + c²-3 ab)

= (a+b+c) (a² + b² + c² - ab - bc - ac).

LHS = RHS, proved.