An algebra problem: Let $a+b+c=0$. Prove that $a^2+b^2+c^2 =6/5$ Let $a,b,c$ be nonzero real numbers such that $a+b+c=0$ and $a^3+b^3+c^3 = a^5 +b^5 +c^5$.  Prove that $a^2+b^2+c^2 =6/5$
I tried to expand $(a+b+c)^5$ but I can't get term of $a^2+b^2+c^2$. 

When you see symmetric equations, try the following substitution $$a+b+c=s=0\qquad ab+bc+ac=q\qquad abc=p$$ And rewrite the equations in terms of $s,q,p$
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)\implies a^3+b^3+c^3=3p$$
$$a^5+b^5+c^5\stackrel{(*)}=-5(ab+bc+ac)(abc)=-5qp$$
(*) means: in this case.
If you're wondering how I obtained this, use Newton's Identities. It isn't hard, but long. 
Finally $$a^3+b^3+c^3=a^5+b^5+c^5\implies -5qp=3p\implies q=-\frac35$$ Now $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=s^2-2q=\frac65$$
