Question: Let $ a, b, c $ be real numbers such that $ a + b + c = 0 $. Prove that $ a^3 + b^3 + c^3 = 3abc $.

Fri, 01 May 2026 07:41:15 +0000

Proving $ a^3 + b^3 + c^3 = 3abc $ Given $ a + b + c = 0 $ Introduction
In algebra, one of the most elegant identities involving symmetric sums of three variables $ a, b, c $ is the identity:
$$
a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)
$$
This identity holds for all real numbers $ a, b, c $. However, a particularly useful special case arises when $ a + b + c = 0 $. In this article, we will prove that if $ a + b + c = 0 $, then
$$
a^3 + b^3 + c^3 = 3abc
$$
through direct algebraic manipulation, using the given condition to simplify the expression. Step-by-Step Proof Start with the well-known algebraic identity valid for any real numbers $ a, b, c $:
$$
a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)
$$
This identity can be verified by expanding the right-hand side, but for now, we accept it as a standard result.