$a^3 + b^3 + c^3 = 3abc$ , can this be true only when $a+b+c = 0$ or $a=b=c$, or can it be true in any other case?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $a=b=c$

Answer 4

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: