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"?
 A: Substituting $$c=-a-b$$ in $$\frac{(a^2+b^2+c^2)(a^3+b^3+c^3)}{6}$$ we obtain $$-(a^2+ab+b^2)ab(a+b)$$ and so is the right-hand side
A: 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.
A: $$a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = -2(ab + bc + ca)$$
$$a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) + 3abc = 3abc$$
$$\begin{align*} a^5 + b^5 + c^5 & = (a + b + c)(a^4 + b^4 + c^4) - (ab + bc + ca)(a^3 + b^2 + c^3) + abc(a^2 + b^2 + c^2) = \\ & = -3abc(ab + bc + ca) - 2abc(ab + bc + ca) = -5abc(ab + bc + ca) \end{align*}$$
So both sides equal $-abc(ab + bc + ca)$.
A: $M= (a^2 + b^2 + c^2)(a^3 + b^3+c^3)=$
$a^5 + b^5 + c^5 + a^3b^2 +a^3c^2 +a^2b^3 + b^3c^2 + a^2c^3 +b^2c^3$
$= (a^5 + b^5 + c^5) + L$ where
$L = a^3b^2 +a^3c^2 +a^2b^3 + b^3c^2 + a^2c^3 +b^2c^3$
$= a(a^2b^2 + a^2c^2) + b(a^2b^2 + b^2c^2) + c(a^2c^2 + b^2b^2)$
$= (a+b+c)(a^2b^2 + a^2c^2 + b^2c^2) - ab^2c^2 -a^2bc^2 - a^2b^2c$
$= - ab^2c^2 -a^2bc^2 - a^2b^2c= -abc(ab + ac + bc)$
On the other hand:
$a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+ac+bc)=-2(ab+ac + bc)$
$= \frac {2L}{abc}$
while $0=(a+b+c)^3 = (a+b)^3 + 3(a+b)^2c + 3(a+b)c^2 + c^3$
$= a^3 + 3a^2b + 3ab^2 + b^3 +3a^2c + 6abc + 3b^2c + 3ac^2 + 3bc^2 + c^3$
$= a^3 + b^3 + c^3 + 3a(ab + ac) +3b(ab + bc) + 3c(ac+ bc) + 6abc$
$=a^3+b^3+c^3 +3(a+b+c)(ab + ac + bc) - 9abc+6abc$
$=a^3 + b^3 +c^3 -3abc$
So $a^3 + b^3+c^3 = 3abc$.
So $M =(a^2+b^2 + c^2)(a^3 + b^3 + c^3) = \frac {2L}{abc}*3abc = 6L$
So $M = a^5 + b^5 + c^5 + L$ and $M = 6L$ so $5L=a^5 + b^5 +c^5$.
And we have:
$\frac {a^2+b^2+c^2}2\times\frac {a^3+b^3 + c^3}3 = \frac M6 = L = \frac{a^5+b^5 + c^5}5$
Yes.  Long and tedious and little insight.  But methodical.
A: 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
$$\begin{align}
6P_5=5P_2P_3\\
\end{align}$$
