`a^3 + b^3 + c^3 = 3abc`
`<=> (a^3 + b^3) +c^3 - 3abc = 0`
`<=> (a+b)^3 - 3ab ( a+b) + c^3 - 3abc = 0`
`<=> (a+b + c)^3 - 3(a+b + c)(a+b+c) - 3ab(a+b+c) = 0`
`<=> (a+b+c)[(a+b+c)^2 - 3c(a+b) - 3ab] = 0`
`<=> (a+b+c)(a^2 + b^2 + c^2 +2ab + 2ac + 2bc -3ab - 3bc-3ab) = 0`
`<=> (a+b+c)(a^2 + b^2 + c^2 - ab -a c - bc) = 0`
`<=>` \(\left[ \begin{array}{l}a+b+c = 0\\2a^2 + 2b^2 + 2c^2 - 2ab - ac - 2bc = 0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}a+b + c = 0\\(a-b)^2 + (a-c)^2 + (b-c)^2 = 0\end{array} \right.\)
Vì `(a-b) ^2 ≥ 0 ∀ a,b ; (a-c)^2 ≥ 0 ∀ a,c ; (b-c)^2 ≥ 0 ∀ b,c`
`=> a - b = 0 ; a - c = 0`
`=> a = b, a= c`
`=> a = b = c`
Vậy..
