# Algebra

• Titu Andreescu
• Bogdan Enescu
Chapter

## Abstract

A very useful algebraic identity is derived by considering the following problem.

Problem 1.1 Factor a 3+b 3+c 3−3abc.

Solution Let P denote the polynomial with roots a,b,c:
$$P(X)=X^{3}-(a+b+c)X^{2}+(ab+bc+ca)X-abc.$$
Because a,b, c satisfy the equation P(x)=0, we obtain
$$\begin{array}{*{20}{c}} {{a^3} - \left( {a + b + c} \right){a^2} + \left( {ab + bc + ca} \right)a - abc = 0,} \\ {{b^3} - \left( {a + b + c} \right){b^2} + \left( {ab + bc + ca} \right)b - abc = 0,} \\ {{c^3} - \left( {a + b + c} \right){c^2} + \left( {ab + bc + ca} \right)c - abc = 0.} \end{array}$$
Adding up these three equalities yields
$$a^{3}+b^{3}+c^{3}- (a+b+c )\bigl(a^{2}+b^{2}+c^{2}\bigr)+(ab+bc+ca)(a+b+c)-3abc=0.$$
Hence Note that the above identity leads to the following result: if a+b+c=0, then a 3+b 3+c 3=3abc.

## Keywords

Algebraic Identities Rearrangement Inequality Well-known Cauchy-Schwarz Inequality Proposed Problem Nonzero Real Number
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.