Proving $2((a+b)^4+(a+c)^4+(b+c)^4)+4(a^4+b^4+c^4+(a+b+c)^4)=3(a^2+b^2+c^2+(a+b+c)^2)^2$ in another way? How do I prove the following identity without expanding both sides directly.
$$2((a+b)^4+(a+c)^4+(b+c)^4)+4(a^4+b^4+c^4+(a+b+c)^4)\\=3(a^2+b^2+c^2+(a+b+c)^2)^2$$
I expanded both sides directly and it is true. However, I was hoping there could be another way to prove it, like that of Candido's identity which could be proved using diagrams.
 A: Why do this by hand?
Mathematica:
Simplify[2 ((a + b)^4 + (a + c)^4 + (b + c)^4) +  4 (a^4 + b^4 + c^4 + (a + b + c)^4) ==  3 (a^2 + b^2 + c^2 + (a + b + c)^2)^2]
(* True *)
A: You are trying to prove that $F=G$ where $F$ and $G$ are symmetric homogeneous 4th degree polynomials in three variables. Fully expanded, such a polynomial is in the form $$A\left(a^4+b^4+c^4\right)+B\left(a^3(b+c)+b^3(a+c)+c^3(a+b)\right)\\+C\left(a^2bc+b^2ac+c^2ab\right)+D\left(a^2b^2+a^2c^2+b^2c^2\right)$$
Imagine the above equals your $F$, and you do not yet know $A,B,C,D$. Substitute four points of the form $(a,b,c)$. You would get four linear equations in $A,B,C,D$. As long as you are not very unlucky, these four equations are independent and linear algebra will tell you the values of $A,B,C,D$.
But the same could be said with $G$ on the right side. And if your solutions for $A,B,C,D$ match, then you may conclude $F=G$.
Really, as long as you trust that you pick four triples $(a,b,c)$ that lead to independent equations, as long as you verify that your left side equals your right side at those four triples, it establishes that $F=G$.
