<s>[INST] 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 $n$?
 [/INST] 
A generalisation for non-negative integer $n$ is provided in this answer. The expansion there can be written as
\begin{align*}
&\,\,\color{blue}{(a+b)^n-\left(a^n+b^n\right)}\\
&\quad\,\,\color{blue}{=ab(a+b)\sum_{k=1}^{\lfloor n/2\rfloor}\left(\binom{n-k}{k}+\binom{n-k-1}{k-1}\right)(-ab)^{k-1}(a+b)^{n-2k-1}}\tag{1}\\
\end{align*}
where we use the binomial identity
$
2\binom{n-k}{k}-\binom{n-k-1}{k}=\binom{n-k}{k}+\binom{n-k-1}{k-1}
$.

We check formula (1) by taking $n=7$. We obtain
\begin{align*}
&\color{blue}{(a+b)^7-\left(a^7+b^7\right)}\\
&\qquad\quad=ab(a+b)\sum_{k=1}^{3}\left(\binom{7-k}{k}+\binom{6-k}{k-1}\right)(-ab)^{k-1}(a+b)^{6-2k}\\
&\qquad\quad=ab(a+b)\left(\left(\binom{6}{1}+\binom{5}{0}\right)(a+b)^4
-\left(\binom{5}{2}+\binom{4}{1}\right)ab(a+b)^2\right.\\
&\qquad\qquad\qquad\qquad\quad\left.+\left(\binom{4}{3}+\binom{3}{2}\right)(ab)^2\right)\\
&\qquad\quad=ab(a+b)\left(7(a+b)^4-14(ab)(a+b)^2+7(ab)^2\right)\\
&\qquad\quad=7ab(a+b)\left((a+b)^4-2(ab)(a+b)^2+(ab)^2\right)\\
&\qquad\quad=7ab(a+b)\left((a+b)^2-ab\right)^2\\
&\qquad\quad\,\,\color{blue}{=7ab(a+b)\left(a^2+ab+b^2\right)^2}
\end{align*}
in accordance with OPs derivation.
</s>