$a+b+c=0;a^2+b^2+c^2=1$ then $a^4+b^4+c^4$ is equal to what? 
$a+b+c=0;a^2+b^2+c^2=1$ then $a^4+b^4+c^4$ is equal to what?

I tried to solve this problem, and I get $a^4+b^4+c^4 = 2(a^2b^2 + 1)$ but I'm not sure if it's  correct

Given $a+b+c = 0$ and $(a^2+b^2+c^2) = 1$, Now $(a^2+b^2+c^2)^2 = 1^2 = 1$
$(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+c^2a^2) = 1.................(1)$
and from $(a+b+c)^2 = 0$, 
we get $\displaystyle 1+2(ab+bc+ca) = 0\Rightarrow (ab+bc+ca) = -\frac{1}{2}$
again squaring both side , we get $\displaystyle (ab+bc+ca)^2  = \frac{1}{4}$
$\displaystyle (a^2b^2+b^2c^2+c^2a^2)+2abc(a+b+c) = \frac{1}{4}\Rightarrow (a^2b^2+b^2c^2+c^2a^2) = \frac{1}{4}$
So put in eqn.... $(1)$ , we get 
$\displaystyle (a^4+b^4+c^4)+2\cdot \frac{1}{4} = 1\Rightarrow (a^4+b^4+c^4) = \frac{1}{2}$
