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

By BYJU'S Exam Prep

Updated on: September 25th, 2023

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)

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1. Prove that (a + b + c)3 – a3 – b3 – c3 = 3 (a + b) (b + c) (c + a)

Adding the common terms:

= (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.

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