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 $n=9$ we have
$$
(a+b)^9-(a^9+b^9)=3(3a^6 + 9a^5b + 19a^4b^2 + 23a^3b^3 + 19a^2b^4 + 9ab^5 + 3b^6)(a + b)ab,
$$
where the first part is irreducible.
