How can we generalize factorisation of $(a+b)^n-(a^n+b^n)$ 
How can we generalize factorisation of
$$(a+b)^n-(a^n+b^n)\,?$$
where $n$ is an odd positive integer.

I found the following cases:
$$(a+b)^3-a^3-b^3=3ab(a+b)$$
$$(a+b)^5-a^5-b^5=5 a b (a + b) (a^2 + a b + b^2)$$
$$(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$$
Exapansions and factoring seem terrible. Maybe there is a good way for generalization.
The question is, can we generalize these factorisations?
$$(a+b)^n-(a^n
+b^n)$$
for all odd integer $n$?

For any case obviously you have $a$ and $b$ in each term since you've removed both the $a$-only and the $b$-only terms from $(a+b)^n$. Also $a=-b$ is always a root, so you'll always get an $(a+b)$ term. So $ab(a+b)$ is a given for any odd $n$.
The roots of $a^2+ab+b^2$ are $a={1\over2}(-b\pm{\sqrt{b^2-4b^2}})=-b\omega,-b\bar\omega$ where $\omega$ is a complex number such that $\omega^3=1$, so if $n=5$ then if $a=-\omega b$ then $$a^5+b^5=b^5(1-\omega^5)=b^5(1-\omega^2)$$ and $$(a+b)^5=b^5(1-\omega)^5=b^5{\bar\omega}^5=b^5\bar\omega^2$$ since $$1-\omega=\bar\omega,\tag{1}$$
and
$$\omega^2=-\bar\omega.\tag{2}$$
You'll get a similar result if $n$ is coprime with $3$, but you don't get that root for $9$. So for example if $n=11$ you get
$$
11ab(a+b)(a^2+ab+b^2)(a^6+3a^5b+7a^4b^2+9a^3b^3+7a^2b^4+3ab^5+b^6)$$
and if $n=13$ you get
$$
13ab(a+b)(a^2+ab+b^2)^2(a^6+3a^5b+8a^4b^2+11a^3b^3+8a^2b^4+3ab^5+b^6)
$$
(obtained using Wolfram Alpha).
In general, using (1),
$$(b-\omega b)^n=b^n(1-\omega)^n=b^n(\bar\omega)^r
$$
where $n=3q+r$, $0<r<3$, and $q,r\in\mathbb{N}$, and
$$
b^n+(-\omega b)^n=b^n(1-\omega^r)=b^n(\bar\omega)^r
$$
if $r=1,2$, so
$$
(b-\omega b)^n-(b^n+(-\omega b)^n)=0
$$
and $a=-\omega b$ is a solution if $r=1,2$ and hence $a^2+ab+b^2$ is a factor.
Since the complex conjugate of a complex root of a real equation is always a root too, then obviously $a=-\bar\omega b$ is also a root.
