Với \({a^3} + {b^3} + {c^3} = 3abc\) thì

Từ đẳng thức đã cho suy ra \({a^3} + {b^3} + {c^3} - 3abc = 0\)

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

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

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

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

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

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

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

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

Do đó nếu \({a^3} + {b^3} + {c^3} - 3abc = 0\) thì \(a + b + c = 0\) hoặc \({a^2} + {b^2} + {c^2} - ab - bc - ac = 0\)

Mà \({a^2} + {b^2} + {c^2} - ab - bc - ac \)\(= \dfrac{1}{2}.\left[ {{{\left( {a - b} \right)}^2} + {{\left( {a - c} \right)}^2} + {{\left( {b - c} \right)}^2}} \right]\)

Nên \({\left( {a - b} \right)^2} + {\left( {a - c} \right)^2} + {\left( {b - c} \right)^2} = 0 \)\(\Leftrightarrow \left\{ \begin{array}{l}a - b = 0\\a - c = 0\\b - c = 0\end{array} \right.\)  suy ra \(a = b = c\).

Vậy \({a^3} + {b^3} + {c^3} = 3abc\) thì \(a = b = c\) hoặc \(a + b + c = 0\).