It is obvious that for odd $n \in \Bbb N$, $a^n-b^n-(a-b)^n$ is divisible by $ab(a-b)$ (with $n=1$ being a special case in which $a^n-b^n-(a-b)^n$ is zero). This can be viewed as a fact about integers, but is more strongly a fact about these polynomials. 


It is apparently also the case that for prime $p > 3$, it is also divisible by $p$ and there is another simple factor
$$
\left. (a^2-ab+b^2) \right| a^p-b^p-(a-b)^p
$$
and in particular, if $p = 6k-1$ for some $k\in\Bbb N$ then $ a^p-b^p-(a-b)^p$ is divisible by $(a^2-ab+b^2)$ but not by $(a^2-ab+b^2)^2$; while if $p = 6k+1$ then $ a^p-b^p-(a-b)^p$ is divisible by $(a^2-ab+b^2)^2$.


This property does not depend on $p$ being prime: The remaining part of the statement is that if odd $n$ is divisible by $3$ then $ a^n-b^n-(a-b)^n$ is not divisible by $(a^2-ab+b^2)$.


So I want to prove that for all $k\in \Bbb N$



> 
> $$
>  a^{6k+1}-b^{6k+1}-(a-b)^{6k+1} = (6k+1) a b (a-b) (a^2-ab+b^2)^2 P\_1(a,b)
> $$
>  for some polynomial $P\_1(a,b) \in \Bbb Z[a,b]$, and 
>  $$
>  a^{6k+5}-b^{6k+5}-(a-b)^{6k+5} = (6k+5) a b (a-b) (a^2-ab+b^2) P\_5(a,b)
> $$
>  for some polynomial $P\_5(a,b)\in \Bbb Z[a,b]$ which itself is not a multiple of $(a^2-ab+b^2)$, and 
>  $$
>  a^{6k+3}-b^{6k+3}-(a-b)^{6k+3} = (6k+3) a b (a-b) P\_3(a,b)
> $$ where $P\_3(a,b)$ is not divisible by $(a^2-ab+b^2)$.
> 
> 
> 


In trying to prove this, I started by noting that $ (a^2-ab+b^2) = (a-\omega b) (a - \omega^2b) $ where $\omega$ is a non-trivial cube root of unity. I was hoping that this would shed some light on why multiples of $3$ would not have the divisibility, but I can't see how this follows.


Is there some more powerful technique that will make the indicated statements "fall out" or at least be easier to prove?



Proving largest power of $(a^2-ab+b^2)$ that divides $a^p-b^p-(a-b)^p$ for odd prime $n$
$\omega$ in what follows is the sixth root of unity.


As user44191 suggested, lets reduce this question to the 1-variable. For the first statement it is enough to prove that


$x^{6k+1} - 1 - (x-1)^{6k+1}$


has a root $\omega$ of order 2. It is clear that $\omega^6 = (\omega-1)^6 = 1$ (because $\omega-1$ is the third root of unity. From this follows


$\omega^{6k+1} - 1 - (\omega-1)^{6k+1} = 0$


The same proof works for $6k+5$, and also closes $6k+3$ case (shows that $\omega$ is not a root in this case)


Now we need to check what happens with the derivative of this polynomial.


For 6k+1 case:


$(6k+1) \omega^{6k} - (6k+1)(\omega - 1)^{6k} = -1$. Root (hence, $\omega$ is a root of order $2$)


For 6k+5 case:


$(6k+5) \omega^{6k+4} - 1 - (6k+5)(\omega - 1)^{6k+4} = (6k+5)(-\omega) - (6k+5)(\omega-1) = 6k+5$. Not a root.


