$a+b+c=0;a^2+b^2+c^2=1$ 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=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
