<s>[INST] 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.
 [/INST] 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.
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