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

Lời giải:

a³ + b³ + c³ = 3 abc

a³ + b³ + c³ – 3 abc = 0

a³ + 3a²b + 3ab² + b³ – (3a²b + 3ab²) + c³ – 3 abc = 0

(a – b)³ + c³ – 3ab(a + b + c) = 0

\(\left( {a + b + c} \right)\left[ {{{\left( {a + b} \right)}^2} - \left( {a + b} \right)c + {c^2}} \right] - 3ab(a + b + c) = 0\)

\(\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} + 2ab - ac - bc} \right) - 3ab(a + b + c) = 0\)

\(\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} + 2ab - ac - bc - 3ab} \right) = 0\)

\(\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - ac - bc} \right) = 0\)

\(\left( {a + b + c} \right)\left( {2{a^2} + 2{b^2} + 2{c^2} - 2ab - 2ac - 2bc} \right) = 0\)

\(\left( {a + b + c} \right)\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - b} \right)}^2}} \right] = 0\)

\(a + b + c = 0\) hoặc \({\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - b} \right)^2} = 0\)

\(a + b + c = 0\) hoặc \(a = b = c\).