<s>[INST] 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 $a^2+b^2+c^2$. 
 [/INST] Hint: With $$c=-a-b$$ we get in $$a^3+b^3+c^3-a^5-b^5-c^5=0$$ the equation
$$ab(a+b)(5a^2+5ab+5b^2-3)=0$$
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