Chứng minh rằng nếu a³ + b³ + c³ = 3abc thì a + b + c = 0 hoặc a = b = c.

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

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

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

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

Với a³ + b³ + c³ = 3abc nên suy ra:

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

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

\[ \Leftrightarrow \left[ \begin{array}{l}a + b + c = 0\\{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = 3ab + 3bc + 3ca\end{array} \right.\]

\[ \Leftrightarrow \left[ \begin{array}{l}a + b + c = 0\\{a^2} + {b^2} + {c^2} - ab - bc - ca = 0\end{array} \right.\]

\[ \Leftrightarrow \left[ \begin{array}{l}a + b + c = 0\\\frac{1}{2}\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] = 0\end{array} \right.\]

\[ \Leftrightarrow \left[ \begin{array}{l}a + b + c = 0\\a = b = c = 0\end{array} \right.\]

Vậy nếu a³ + b³ + c³ = 3abc thì a + b + c = 0 hoặc a = b = c.