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.
 A: Hint: $(a-b)^4=a^4-4a^3b+6a^2b^2-4ab^3+b^4$
A: 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.
A: $$(a-b)^4=(a^2+b^2-2ab)^2=(a^2+b^2)^2+4a^2b^2-4ab(a^2+b^2)$$
$$a^4+b^4=(a^2+b^2)^2-2a^2b^2$$
On addition, $$2((a^2+b^2)^2+a^2b^2-2ab(a^2+b^2))=2(a^2+b^2-ab)^2$$
A: We begin with the special case that $a=x$ and $b=1$ to obtain $f(x)=x^4 +(x-1)^4 =2g(x),$ where $$g(x) =x^4-2x^3+3x^2-2x+1.$$ Since $g(x)$ has no real roots, it equals the product of two irreducible quadratic polynomials. Let us guess that these two  quadratic polynomials are equal, so that $g(x)=(x^2+cx+1)^2.$ Happily, $c=-1$ works. Therefore,
$$f(x)=2(x^2-x+1)^2.$$
In the above factorization replace $x$ by $a\over b$ and multiply both sides by $b^4$ to get the factorization of the original expression.
