Prove $a^2+b^2+c^2=\frac{6}{5}$ if $a+b+c=0$ and $a^3+b^3+c^3=a^5+b^5+c^5$ 
if $a,b,c$ are real numbers that $a\neq0,b\neq0,c\neq0$ and $a+b+c=0$ and $$a^3+b^3+c^3=a^5+b^5+c^5$$
Prove that $a^2+b^2+c^2=\frac{6}{5}$.

Things I have done: $a+b+c=0$ So $$a^3+b^3+c^3=a^5+b^5+c^5=3abc$$
also $$a^2+b^2+c^2=-2ab-2bc-2ca$$
I tried multiplying $(a^3+b^3+c^3)$ and $a^2+b^2+c^2$ but I was not able to reach a useful result from that.

There are probably cleverer methods for this problem, but here is an approach using the general theory of what are called symmetric polynomials. We use the Newton-Girard identities in order to write power-sum symmetric polynomials in terms of elementary symmetric polynomials. Here,
$$p_k:=x_1^k+\cdots+x_n^k \qquad e_k=\sum_{1\le i_1<\cdots<i_k\le n} x_{i_1}\cdots x_{i_k}$$
In particular, with three variables $a,b,c$ we have
$$p_1=a+b+c, \quad p_2=a^2+b^2+c^2, \quad p_3=a^3+b^3+c^3, \quad p_5=a^5+b^5+c^5 $$
$$e_1=a+b+c, \quad e_2=ab+bc+ac, \quad e_3=abc.$$
We know that $p_1=e_1$ which in our case $=0$, and that $p_3=p_5$. Using the formulas for $p_k$ in terms of the other $e_k$s given in the link, and setting $e_1=0$, we have
$$p_2=-2e_2, \quad p_3=3e_3, \quad p_5=-5e_3e_2.$$
Since $p_3=p_5\implies 3e_3=-5e_3e_2\implies e_2=-\frac{3}{5}$, we can plug into $p_2=-2e_2$ to get $p_2=\frac{6}{5}$.
