Chứng minh ${a^3} + {b^3} + {c^3} = 3abc$.

Ta có:

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

$= {\left( {a + b} \right)^3} - 3ab\left( {a + b} \right) + {c^3} - 3abc = \left[ {{{\left( {a + b} \right)}^3} + {c^3}} \right] - \left[ {3ab\left( {a + b} \right) - 3abc} \right]$

$= \left( {a + b + c} \right)\left[ {{{\left( {a + b} \right)}^2} - \left( {a + b} \right)c + {c^2}} \right] - 3ab\left( {a + b + c} \right)$

$= \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$

Mà a + b + c = 0

$\Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 0$

$\Rightarrow {a^3} + {b^3} + {c^3} = 3abc$ (đpcm).