Algebra

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.