Is it always possible to factorize $(a+b)^p - a^p - b^p$ this way? I'm looking at the solution of an IMO problem and in the solution the author has written the factorization $(a+b)^7 - a^7 - b^7=7ab(a+b)(a^2+ab+b^2)^2$ to solve the problem. It seems like it's always possible to find a factorization like $(a+b)^p - a^p - b^p=p\cdot(ab)^{\alpha_1}(a+b)^{\alpha_2}(a^2+ab+b^2)^{\alpha_3}P_{\omega}(a,b)$ for any prime number p where $P_{\omega}(a,b)$ is an irreducible homogenous polynomial of degree $\omega$ and $\alpha_1 + \alpha_2 + \alpha_3 + \omega = p-1$. I wonder if it's true in general.
Here are some examples:
$(a+b)^2 - a^2 - b^2 = 2ab$
$(a+b)^3 - a^3 - b^3 = 3ab(a+b)$
$(a+b)^5 - a^5 - b^5 = 5ab(a+b)(a^2+ab+b^2)$
$(a+b)^7 - a^7 - b^7 = 7ab(a+b)(a^2+ab+b^2)^2$
$(a+b)^{11}- a^{11} - b^{11} = 11 ab(a + b)(a^2 + a b + b^2)(a^6 + 3 a^5 b + 7 a^4 b^2 + 9 a^3 b^3 + 7 a^2 b^4 + 3 a b^5 + b^6)$
I've proved the following statements:
For any prime number p:
$p|(a+b)^p - a^p-b^p$
$ab|(a+b)^p - a^p-b^p$
For any odd prime number:
$(a+b) | (a+b)^p - a^p-b^p$ because $(a+b)^p - a^p-b^p$ vanishes when $a=-b$ if p is odd.
For any prime number $p \geq 5$:
$ (a^2 + ab + b^2) | (a+b)^p - a^p - b^p$
Proof: Any prime number greater than 3 is of the form $p=6k+1$ or $p=6k+5$. On the other hand, if we set $\Large \omega= e^\frac{2\pi i}{3}$ we have: $1 + \omega + \omega^2 = 0$, therefore 
$(1+\omega)^3 = 1 + 3\omega + 3\omega^2 + \omega^3 = 2 + 3(\omega+\omega^2) = -1$
$(1+\omega^2)^3 = 1 + 3\omega^2 + 3\omega^4 + \omega^3 = 2 + 3(\omega^2+\omega) = -1$
We have $a^2+ab+b^2 = (a-b\omega)(a-b\omega^2)$ in $\mathbb{Z}[\omega]$, I'll show that both $a-b\omega$ and $a-b\omega^2$ divide $(a+b)^p - a^p - b^p$ for every $p>3$:
if $p=6k+1$ and we replace $a=b\omega$ then:
$(a+b)^p=(b\omega+b)^{6k+1} = b^{6k+1} (1+\omega)^{6k} (1+\omega) = b^{6k+1}(1+\omega)$
$(a+b)^p - a^p - b^p = b^{6k+1}(1+\omega) - (b\omega)^{6k+1} - b^{6k+1} = b^{6k+1}(1+\omega) - b^{6k+1}(\omega+1) = 0$
If $p=6k+5$ and we replace $a=b\omega$ then:
$(a+b)^p=(b\omega+b)^{6k+5} = b^{6k+5} (1+\omega)^{6k} (1+\omega)^5 = (-1)b^{6k+1}(1+\omega)^2$
$(a+b)^p - a^p - b^p = (-1)b^{6k+5}(1+\omega)^2 - (b\omega)^{6k+5} - b^{6k+5} = (-1)b^{6k+5}(1+\omega)^2 - b^{6k+5}(\omega^2+1) = p^{6k+5}(-1-2\omega-\omega^2-\omega^2-1)=0$
The same could be shown for $a=b\omega^2$.
Therefore both $(a-b\omega)$ and $(a-b\omega^2)$ are factors of $(a+b)^p-a^p-b^p$ but I'm not sure if that is sufficient to conclude $a^2+ab+b^2=(a-b\omega)(a-b\omega^2) | (a+b)^p - a^p - b^p$
So, I guess I have proved that $p$, $ab$, $a+b$ and $a^2+ab+b^2$ all divide $(a+b)^p-a^p-b^p$, but I have no idea how to show that $P_{\omega}(a,b)$ must be irreducible. I have verified this conjecture up to $p=97$ and it's true I think. Any ideas on how to prove that?

The polynomials $P_n(x)=P_n(x,1)$ are called the Cauchy-Mirimanoff polynomials.
They are defined for all $n$ natural $n\geq 2$ by:
$$ (X^n+1)-X^n-1 = X(X+1)^E(X^2+X+1)^eP_n(X) $$
Where $E=0$ if n is even, and 1 otherwise, and, as @KierenMacMillan notice, $e= 0,1,2 $ according to $n= 0,2,1\mod 3$.
We know that $P_{2p}(x)$ and $P_{3p}(x)$ are irreducibles when $p$ is prime. It's still unknown whether all Cauchy-Mirimanoff polynomials are irreducibles...
For example, the proof that $P_{3p}(x)$ is irr, use the fact that those polynomials are self-reciprocals $(a_0=a_n, a_1=a_{n-1}, a_2=a_{n-2}, ...)$, and the degree of $P_n$ is even; this ensures the existence of a unique polynomial $P_n^*$ called the "reciprocal transform" of $P_n$, concretely, if $deg(P_n)=2d$:
$$P_n(x) = x^{d} P_n^*\Big(x+\frac{1}{x}\Big)$$
The irreducibility of $P^*$ and $P$ are directly linked through Dikson's Theorem, which states (in the case of the Cauchy-Mirimanoff polynomials, with even degree) that $P$ is irreducible $\Leftrightarrow$ $P^*$ is irreducible.
Finally, Netwon polygons are used to proof that $P^*$ is irreducible.
The details can be found here:
https://trace.tennessee.edu/utk_graddiss/707/
