$a+b+c=0;a^2+b^2+c^2=1$ 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
 A: 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 
$\displaystyle (a^4+b^4+c^4)+2\cdot \frac{1}{4} = 1\Rightarrow (a^4+b^4+c^4) = \frac{1}{2}$
A: $$a+b+c=0\Rightarrow c=-(a+b)$$
$$1=a^2+b^2+(a+b)^2=2a^2+2ab+2b^2=2(a^2+ab+b^2)\Rightarrow a^2+ab+b^2=\frac12$$
$$\begin{align*}
a^4+b^4+c^4 &= a^4+b^4+(a^4+4a^3b+6a^2b^2+4ab^3+b^4)\\
&= 2(a^4+2a^3b+3a^2b^2+2ab^3+b^4)\\
&= 2(a^2(a^2+ab+b^2)+b^2(a^2+ab+b^2)+ab(a^2+ab+b^2))\\
&= a^2+ab+b^2=\frac12
\end{align*}$$
A: Given a+b+c=0a+b+c=0 and (a2+b2+c2)=1(a2+b2+c2)=1, Now (a2+b2+c2)2=12=1(a2+b2+c2)2=12=1
(a4+b4+c4)+2(a2b2+b2c2+c2a2)=1.................(1)(a4+b4+c4)+2(a2b2+b2c2+c2a2)=1.................(1)
and from (a+b+c)2=0(a+b+c)2=0,
we get 1+2(ab+bc+ca)=0⇒(ab+bc+ca)=−121+2(ab+bc+ca)=0⇒(ab+bc+ca)=−12
again squaring both side , we get (ab+bc+ca)2=14(ab+bc+ca)2=14
(a2b2+b2c2+c2a2)+2abc(a+b+c)=14⇒(a2b2+b2c2+c2a2)=14(a2b2+b2c2+c2a2)+2abc(a+b+c)=14⇒(a2b2+b2c2+c2a2)=14
So put in eqn.... (1)(1) , we get
(a4+b4+c4)+2⋅14=1⇒(a4+b4+c4)=12
