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"?

The theory for this type of questions are Newton's Identities.
For a cubic polynomial
$$\begin{align}ax^3+bx^2+cx+d\end{align}\\$$
Vieta's formulas for the roots $\alpha_1, \alpha_2, \alpha_3$ are:
$$\begin{align}a(\alpha_1 + \alpha_2 + \alpha_3) + b=0\tag{V}\\
a(\alpha_1\alpha_2 + \alpha_2\alpha_3 + \alpha_3\alpha_1) - c=0\\
a\alpha_1\alpha_2\alpha_3 + d=0\end{align}$$
Newton's Identities for the power sums $P_i:=\alpha_1^i+\alpha_2^i+\alpha_3^i$ are:
$$\begin{align}
aP_1 ~+ ~&b&=0\tag{N1}\\
aP_2 ~+ ~&bP_1 +~ 2c&=0\tag{N2}\\
aP_3 ~+ ~&bP_2 +~ cP_1 +~ 3d&=0\tag{N3}\\
aP_4 ~+ ~&bP_3 +~ cP_2 +~ dP_1&=0\tag{N4}\\
aP_5 ~+ ~&bP_4 +~ cP_3 +~ dP_2&=0\tag{N5}\\
\end{align}$$
If $\alpha_1 + \alpha_2 + \alpha_3=0$ then from $(V)$ it follows that $b=0$. Using $(N5)$
$$\begin{align}
aP_5 +~ cP_3 +~ dP_2&=0\\
\end{align}$$
multiply by $6$
$$\begin{align}
6aP_5 +~ 3\cdot2cP_3 +~ 2\cdot3 dP_2&=0\\
\end{align}$$
using $(N2)$ and $(N3)$
$$\begin{align}
6aP_5 -~ 3\cdot(aP_2+bP_1)P_3 -~ 2\cdot(aP_3+bP_2+cP_1)P_2=\end{align}$$
$$\begin{align}
6aP_5-3aP_2P_3 -~ 2aP_3P_2=0\\
\end{align}$$
If $a\neq 0$ then
$$\begin{align}
6P_5=5P_2P_3\\
\end{align}$$
