# Prove that (a + b + c)3 - a3 - b3 - c3 = 3 (a + b) (b + c) (c + a)

By Ritesh|Updated : November 14th, 2022

We have to prove, (a + b + c)3 - a3 - b3 - c3 = 3 (a + b) (b + c) (c + a)

First consider LHS = [(a + b + c)3 - a3] - (b3 - c3)

Using identity a3 + b3 = (a + b) (a2 + b2 - ab) and a3 - b3 = (a - b) (a2 + b2 + ab) the above equation becomes:

= (a + b + c - a) [(a + b + c)2 + a2 + a (a + b + c)] - [(b + c) (b2 + c2 - bc)]

In simplification we get:

= (b + c) [a2 + b2+ c2 + 2ab + 2bc + 2ca + a2 + a2 + ab + ac] - (b + c) (b2 + c2 - bc)

= (b + c) [b2 + c2 +3a2 + 3ab + 3ac - b2 - c2 + 3bc]

Taking 3 as common we get:

= (b + c) [3 (a2 + ab + ac + bc)]

Taking a and c as common:

= 3 (b + c) [a (a + b) + c (a + b)]

= 3 (b + c) [(a + c) (a + b)]

In simplification we get the:

= 3 (a + b) (b + c) (c + a)

= RHS

Hence proved.

All mathematical simple and complex formulas are built on algebraic equations. The equations in which all possible values for the variables are legitimate and the equation will be true regardless of the value the variables have been known as algebraic identities. Advanced mathematical equations, analysis, and research-based notions all make use of algebraic identities. These equations are crucial for math students to understand since they are essential for resolving engineering and scientific issues.

Summary:

## Prove that (a + b + c)3 - a3 - b3 - c3 = 3 (a + b) (b + c) (c + a)

It is proved that (a + b + c)3 - a3 - b3 - c3 = 3 (a + b) (b + c) (c + a). Algebraic identities are equations in algebra that are true regardless of the value of each of their variables. Mathematical equations with numbers, variables (unknown values), and mathematical operators are known as algebraic identities and expressions (addition, subtraction, multiplication, division, etc.)