If $a+b+c=0$ prove that $ (a^2+b^2+c^2) \times (a^2+b^2+c^2) = 2(a^4+b^4+c^4)$ 
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$(a^2+b^2+c^2) \times (a^2+b^2+c^2) = 2(a^4+b^4+c^4)$$

What is a good way to do this? 
This question came from answering this slightly harder question. Those answers were somewhat hard to understand for me. To get something easier to digest I made a very similar but easier (lower exponent) question.

Let $e_1 = a+b+c, e_2 = ab+bc+ac, e_3 = abc$. 
Since $A = (a^2+b^2+c^2)^2-2(a^4+b^4+c^4)$ is symmetric and homogeneous of degree $4$, it is a linear combination of $e_1^4, e_1^2e_2, e_1e_3, e_2^2$, and when $e_1=0$ the only nonzero one is possibly $e_2^2$ :
There is a coefficient $k$ such that if $e_1=0$ then $A = ke_2^2$.
To check that $k$ is $0$, we only need to check that the identity is true for one case where $e_1=0$ and $e_2 \neq 0$.
Picking for example $a=0,b=1,c=-1$, you get $4=4$ so things work out.
