<s>[INST] If $a+b+c=0$ prove that $ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5} $ 
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5} $$

I've tried squaring, cubing, etc. the $a+b+c=0$, but I've just dug myself in.
Is there a more elegant way to prove this, or a way to find the right "trick"?
 [/INST] Let' s try like this: Suppose $a,b,c$ are the roots of the polynomial $x^3-mx-n$. 
By Vieta's formula's we have:
 $$a^2+b^2+c^2 = (a+b+c)^2-2(ab+bc+ca)= 0-2m = 2m$$
Since $x^3 =mx+n \;\;\;\;(*)$ we have:
$$x^5 = x^2(mx+n) = mx^3+nx^2= m^2x+mn+nx^2$$
so
$$a^5+b^5+c^5 = m^2(a+b+c)+3mn+n(a^2+b^2+c^2) =3mn+n(a^2+b^2+c^2)=5mn$$
Also because of $(*)$ we have $$a^3+b^3+c^3 = m(a+b+c) +3n = 3n$$
and we are done.
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