<s>[INST] 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.
 [/INST] 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 $p_2=\frac{6}{5}$.
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