Factoring $a^4+b^4+(a-b)^4$ I'm trying to factor $$a^4+b^4+(a-b)^4$$ so the result would be $2(a^2-ab+b^2)^2$ but I can't get that.

I rewrite it as:
$$a^4+b^4+(a-b)^4=(a^2+b^2)^2-2a^2b^2+(a-b)^4=(a^2-\sqrt2 ab+b^2)(a^2+\sqrt2 ab+b^2)+(a-b)^4$$
But I can't use difference of squares anymore because $(a-b)^4$ is not negative.

It's a variation of Candido's identities, its proof without words are as follows:

Now
\begin{align}
  [x^2+y^2+(x+y)^2]^2 &= 2[x^4+y^4+(x+y)^4] \\
  [a^2+(-b)^2+(a-b)^2]^2 &= 2[a^4+(-b)^4+(a-b)^4] \\
  [2a^2-2ab+2b^2]^2 &= 2[a^4+b^4+(a-b)^4] \\
  4(a^2-2ab+b^2)^2 &= 2[a^4+b^4+(a-b)^4] \\
  a^4+b^4+(a-b)^4 &= 2(a^2-ab+b^2)^2
\end{align}
Refer to link 1 and link 2 for your further interests.
