<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] One way of looking at this is to let $a,b,c$ be the roots of $x^3-px^2+qx-r=0$
Now $p=a+b+c=0$ so we have $x^3+qx-r=0$ also $q=ab+bc+ca$ and $r=abc$
If we substitute $x=a,b,c$ and add we get $a^3+b^3+c^3+q(a+b+c)-3r=0$ so that $a^3+b^3+c^3=3r=a^5+b^5+c^5$
Now multiply by $x^2$ to give $x^5+qx^3-rx^2=0$ - again substitute $x=a,b,c$ and add$$a^5+b^5+c^5+q(a^3+b^3+c^3)-r(a^2+b^2+c^2)=0$$ so that (substituting the value $3r$ for the sums of odd powers) $$a^2+b^2+c^2=3+3q$$
We also have $0=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=a^2+b^2+c^2+2q$ from which we get $q=-\frac 35$ and $a^2+b^2+c^2=\frac 65$
The main point is to work from the cubic, which saves a lot of work having to remember or reconstruct identities involving symmetric polynomials, and allows for early use of the given conditions. Thus the workings become more straightforward.
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