Factoring $a^4+b^4+(a-b)^4$

Question

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.

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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.

Answer 1

$$(a-b)^4=(a^2+b^2-2ab)^2=(a^2+b^2)^2+4a^2b^2-4ab(a^2+b^2)$$

$$a^4+b^4=(a^2+b^2)^2-2a^2b^2$$

On addition, $$2((a^2+b^2)^2+a^2b^2-2ab(a^2+b^2))=2(a^2+b^2-ab)^2$$

Answer 2

Hint: $(a-b)^4=a^4-4a^3b+6a^2b^2-4ab^3+b^4$

Answer 3

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.

Answer 4

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.