<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] Note that $(a^3+b^3+c^3)-(a+b+c)(a^2+b^2+c^2)+(ab+bc+ca)(a+b+c)-3abc = 0$.
Using the fact that $a+b+c = 0$, this reduces to $a^3+b^3+c^3 = 3abc$. Thus, $a^5+b^5+c^5 = 3abc$.
Now, note that $(a^5+b^5+c^5)-(a+b+c)(a^4+b^4+c^4)+(ab+bc+ca)(a^3+b^3+c^3)-abc(a^2+b^2+c^2) = 0$. 
Using the facts that $a+b+c = 0$ and $a^5+b^5+c^5 = a^3+b^3+c^3 = 3abc$, this reduces to 
$(a^2+b^2+c^2)-3(ab+bc+ca) = 3$ [*]. 
Since $a+b+c = 0$, we have $0 = (a+b+c)^2 = (a^2+b^2+c^2)+2(ab+bc+ca)$. 
Hence, $ab+bc+ca = -\frac{1}{2}(a^2+b^2+c^2)$. Substituting this into [*] gives: 
$(a^2+b^2+c^2)+\frac{3}{2}(a^2+b^2+c^2) = 3$, i.e. $a^2+b^2+c^2 = \frac{6}{5}$, as desired.
Sidenote: Newton's Sums is very convenient for these types of problems.  
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