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.

$$(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)=$$
$$=\sum_{cyc}(2a^2b^2-a^4)=(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0.$$
Because
$$\sum_{cyc}(2a^2b^2-a^4)=4a^2b^2-(a^4+b^4+c^4-2a^2c^2-2b^2c^2+2a^2b^2)=$$
$$=(2ab)^2-(a^2+b^2-c^2)^2=(2ab-a^2-b^2+c^2)(2ab+a^2+b^2-c^2)=$$
$$=(c^2-(a-b)^2)((a+b)^2-c^2)=(c-a+b)(c+a-b)(a+b-c)(a+b+c).$$
