<s>[INST] Factor $(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)$ to get $(b+c-a) (a+b-c) (a-b+c) (a+b+c)$ 
Is there an easy way to factor $(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)$ to get $(b+c-a) (a+b-c) (a-b+c) (a+b+c)$?

I have tried various methods such as completing the square, but I wasn't able to figure it out as it became very messy.
 [/INST] Expand and notice that it's "trying to be" $-(\pm a^2 \pm b^2 \pm c^2)^2$
for some choice of $\pm$ signs.  None of the sign choices quite works, but they're each off by a square, e.g.
$$
(a^2+b^2+c^2)^2-2(a^4+b^4+c^4) = 4a^2b^2 - (a^2+b^2-c^2)^2.
$$
So you can factor it as the difference of two squares, and then each factor is itself a difference of two squares and thus splits further into linear forms, giving the desired factorization.
Note that the factored form is within a constant factor $16$ of Heron's formula for the square of a triangle of side lengths $a,b,c$.
</s>