Với \(a^3 + b^3 + c^3 = 3abc\) thì

Từ đẳng thức đã cho suy ra

\(\begin{array}{l} {a^3} + {b^3} + {c^3} - 3abc = 0\\ {b^3} + {c^3} = \left( {b + c} \right)\left( {{b^2} + {c^2} - bc} \right)\\ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = {a^3} + \left( {{b^3} + {c^3}} \right) - 3abc\\ \Leftrightarrow {a^3} + {b^3} + {c^3} - 3abc = {a^3} + {\left( {b + c} \right)^3} - 3bc\left( {b + c} \right) - 3abc\\ \Leftrightarrow {a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} - a\left( {b + c} \right) + {{\left( {b + c} \right)}^2}} \right) - \left[ {3bc\left( {b + c} \right) + 3abc} \right]\\ \Leftrightarrow {a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} - a\left( {b + c} \right) + {{\left( {b + c} \right)}^2}} \right) - 3bc\left( {a + b + c} \right)\\ \Leftrightarrow {a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} - a\left( {b + c} \right) + {{\left( {b + c} \right)}^2} - 3bc} \right) \Leftrightarrow {a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} - ab - ac + {b^2} + 2bc + {c^2} - 3bc} \right)\\ \Leftrightarrow {a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - ac - bc} \right) \end{array}\)

Do đó nếu 

\({a^3} + {b^3} + {c^3} - 3abc = 0 \to \left[ \begin{array}{l} a + b + c = 0\\ {a^2} + {b^2} + {c^2} - ab - bc - ac = 0 \end{array} \right.\)

Mà 

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

Nên 

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

Vậy \({a^3} + {b^3} + {c^3} = 3abc \to \left[ \begin{array}{l} a = b = c\\ a + b + c = 0 \end{array} \right.\)