Cho a + b + c = 0. Chứng minh a³ + b³ + c³ = 3abc.

Ta có: \({\left( {a + b + c} \right)^3}\)

\( = {a^3} + {b^3} + {c^3} + 3{a^2}b + 3a{b^2} + 3{b^2}c + 2b{c^2} + 3{c^2}a + 3c{a^2} + 6abc\)

\( = {a^3} + {b^3} + {c^3} + \left( {3{a^2}b + 3a{b^2} + 3abc} \right) + \left( {3{b^2}c + 2b{c^2} + 3abc} \right) + \left( {3{c^2}a + 3c{a^2} + 3abc} \right) - 3abc\)

\( = {a^3} + {b^3} + {c^3} + 3ab\left( {a + b + c} \right) + 3bc\left( {a + b + c} \right) + 3ca\left( {a + b + c} \right) - 3abc\)

Với a + b + c = 0 nên suy ra \({a^3} + {b^3} + {c^3} - 3abc = 0\).

Hay a³ + b³ + c³ = 3abc (đpcm).