$a^3 + b^3 + c^3 = 3abc$ , 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.
 A: Without loss of generality, suppose that $a\ge b$ and $a\ge c$. Then $$a^2+b^2+c^2-ab-bc-ca=(a-b)(a-c)+(b-c)^2\ge 0.$$
Equality can only occur if $a=b=c$ and so there are no other solutions.
A: By your work:
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$
Thus, $$a^3+b^3+c^3-3abc=0$$ for $a+b+c=0$ or for
$$a^2+b^2+c^2-ab-ac-bc=0,$$ which is
$$(a-b)^2+(a-c)^2+(b-c)^2=0,$$ which gives $$a=b=c.$$
Id est, we have no another cases for equality occurring for real values. 
A: The equation can also be factorized as follows-
$$\begin{align}a^3+b^3+c^3-3abc&=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\\
&=\frac 12(a+b+c)(2a^2+2b^2+2c^2-2ab-2bc-2ca)\\
&=\frac 12(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)
\end{align}
$$
Hence the equation can only be true when either $a+b+c=0 $ or $a=b=c$
A: 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:
 
