# 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"?

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

$$a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = -2(ab + bc + ca)$$ $$a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) + 3abc = 3abc$$ \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 $-abc(ab + bc + ca)$.

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.

$M= (a^2 + b^2 + c^2)(a^3 + b^3+c^3)=$

$a^5 + b^5 + c^5 + a^3b^2 +a^3c^2 +a^2b^3 + b^3c^2 + a^2c^3 +b^2c^3$

$= (a^5 + b^5 + c^5) + L$ 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.

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}