Intuition behind problem in (classical) algebra To give some background, the question is to show that if $a=b+c$ then $$a^4+b^4+c^4 = 2a^2b^2+2b^2c^2+2c^2a^2$$ 
Which, for completeness, I was able to do by squaring twice $$(a-b-c)^2=0$$ gives $$a^2+b^2+c^2= 2(bc-ac-ab)$$ which on squaring gives the required identity. $$a^4+b^4+c^4 +2a^2b^2+2b^2c^2+2c^2a^2 = 4a^2b^2+4b^2c^2+4c^2a^2 + 8(a^2bc - ab^2c-abc^2)$$ $$a^4+b^4+c^4 =2a^2b^2+2b^2c^2+2c^2a^2 +8abc(a-b-c) $$
$$a^4+b^4+c^4 = 2a^2b^2+2b^2c^2+2c^2a^2$$ 
For the second part the statement is: "explain why an unsymmetric equation gives such a symmetric expression." I ignored this part while solving it as I did not have a very good impression of the author of the problem set from the previous problems and as no rigorous definition of symmetry was given the author was presumably requiring some kind of aesthetic appreciation. 
The reason as it turns out in the hint is that "on squaring twice effect of minus signs gets cancelled." If this statement is taken for granted, from this logic all the ring permutations of ($\pm a,\pm b,\pm c$),i.e 4 yield the same equation. It is indeed so as the factors turn out to be $$(a+b+c)(a-b-c)(a+b-c)(a-b+c)$$
However, by this same logic $(a+b+c)^4=(a-b+c)^4$ But it is not. The reason I am asking here  is that I can verify the above algebraically and see why everything is ok, but cannot explain it as the way algebra works for me is that I work it out all keeping care for all the minus signs then magically half the terms seem to cancel out and I am left with a single expression out of the blue. 
Sorry for the wall of text.

Expanding the area of the triangle $ABC$ of sides $a,b,c$ by Heron's formula, you'll get 
$$ Area(ABC)=\frac{1}{16} (2b^2c^2+2c^2a^2+2a^2b^2-a^4-b^4-c^4)$$
If $a=b+c$ then the triangle is degenerate and has area zero. Therefore
$$2b^2c^2+2c^2a^2+2a^2b^2-a^4-b^4-c^4=0$$
holds.
