Ta có a3 + b3 + c3 = 3abc
<=> (a + b)3 - 3ab(a + b) + c3 = 3abc
<=> (a + b + c)[(a + b)2 - (a + b)c + c2] - 3ab(a + b + c) = 0
<=> (a + b + c)(a2 + 2ab + b2 - ac - bc + c2 - 3ab) = 0
<=> (a + b + c)(a2 + b2 + c2 - ab - ac - bc) = 0
<=> \(\orbr{\begin{cases}a+b+c=0\left(\text{tmđk}\right)\\a^2+b^2+c^2-ab-ac-bc=0\end{cases}}\)
Khi a2 + b2 + c2 - ab - ac - bc = 0
<=> 2a2 + 2b2 + 2c2 - 2ab - 2ac - 2bc = 0
<=> (a2 - 2ab + b2) + (b2 - 2bc + c2) + (a2 - 2ac + c2) = 0
<=> (a - b)2 + (b - c)2 + (c - a)2 = 0
<=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\left(\text{loại}\right)\)
Vậy a + b + c = 0
