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?