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.

let $$T_{n}=a^n+b^n+c^n,T_{1}=0,T_{3}=T_{5},\Longrightarrow -(ab+bc+ac)=\dfrac{T_{2}}{2}$$
and we have
$$T_{n+2}=(a+b+c)T_{n+1}-(ab+bc+ac)T_{n}+abcT_{n-1}$$
so
$$T_{n+2}=-(ab+bc+ac)T_{n}+abcT_{n-1}$$
so
$$T_{4}=-(ab+bc+ac)T_{2}+abcT_{1}=\dfrac{T^2_{2}}{2}$$
$$T_{5}=-(ab+bc+ac)T_{3}+abc T_{2}=\dfrac{T_{2}T_{3}}{2}+\dfrac{T_{3}T_{2}}{3}$$
since
$$T_{5}=T_{3}$$
so
$$T_{2}=\dfrac{6}{5}$$
