<s>[INST] $a^3 + b^3 + c^3 = 3abc$ , can this be true only when $a+b+c = 0$ or $a=b=c$, or can it be true in any other case? Since 
$$
a^3 + b^3 + c ^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc ),
$$ from this it is clear that if $(a+b+c) = 0$ , then $a^3 + b^3 + c ^3 - 3abc = 0$ and
$a^3 + b^3 + c ^3 = 3abc$ . Also if $a=b=c$ , $a^3 + b^3 + c^3 = a^3 + a^3 +a^3 = 3a^3 = 3aaa = 3abc$ , hence $a^3 + b^3 + c^3 = 3abc$.
But I was wondering if $a^3 + b^3 + c^3 = 3abc$ can be true even when none of the above two relations are true. Please guide me toward a solution.
 [/INST] The polynomial factors fully over the complex numbers,
$$ (a+b+c)(a+b \omega + c \omega^2)(a + b \omega^2 + c \omega) \; , \; $$
where $\omega$ is a primitive cube root of unity, either solution of $x^2 + x + 1 =0$
There is actually a concrete calculation that tells us whether a homogeneous cubic factors completely over the complexes. Here is an excerpt from an article by Brookfield:
 
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