Calculate

a = -(b + c), a^3 = -(b + c)^3 = -b^3 - 3bc(b + c) - c^3, a^3 + b^3 + c^3 = -3bc*(-a) = 3abc. Ed Gray

$\Rightarrow a^3+b^3+c^3=\color{#D61F06}{x}$

Adding $(-3abc)$ both sides.

$\Rightarrow a^3+b^3+c^3-3abc=\color{#D61F06}{x}-3abc$
$\Rightarrow \color{#3D99F6}{(a+b+c)}(a^2+b^2+c^2-ab-bc-ca)=\color{#D61F06}{x}-3abc$
$\Rightarrow \color{#3D99F6}{0}=\color{#D61F06}{x}-3abc$
$\Rightarrow \boxed{a^3+b^3+c^3=3abc}$

We know that: $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$

So, when $(a+b+c)=0$ ;

$a^3+b^3+c^3-3abc=(0)(a^2+b^2+c^2-ab-bc-ac)\\a^3+b^3+c^3-3abc=0\\\therefore\boxed{a^3+b^3+c^3=3abc}$