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.
 A: 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.  
A: There are probably cleverer methods for this problem, but here is an approach using the general theory of what are called symmetric polynomials. We use the Newton-Girard identities in order to write power-sum symmetric polynomials in terms of elementary symmetric polynomials. Here,
$$p_k:=x_1^k+\cdots+x_n^k \qquad e_k=\sum_{1\le i_1<\cdots<i_k\le n} x_{i_1}\cdots x_{i_k}$$
In particular, with three variables $a,b,c$ we have
$$p_1=a+b+c, \quad p_2=a^2+b^2+c^2, \quad p_3=a^3+b^3+c^3, \quad p_5=a^5+b^5+c^5 $$
$$e_1=a+b+c, \quad e_2=ab+bc+ac, \quad e_3=abc.$$
We know that $p_1=e_1$ which in our case $=0$, and that $p_3=p_5$. Using the formulas for $p_k$ in terms of the other $e_k$s given in the link, and setting $e_1=0$, we have
$$p_2=-2e_2, \quad p_3=3e_3, \quad p_5=-5e_3e_2.$$
Since $p_3=p_5\implies 3e_3=-5e_3e_2\implies e_2=-\frac{3}{5}$, we can plug into $p_2=-2e_2$ to get $p_2=\frac{6}{5}$.
A: You should be able to get something by looking at the product $a^3+b^3+c^3$ and $a^2+b^2+c^2$. Note that
$$a^2b^3+a^3b^2=a^2b^2(a+b)=-(ab)(abc).$$
A: 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.
A: 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
$$T_{2}=\dfrac{6}{5}$$
