This problem come across when posting Generalization of $x^3-2=(x-\sqrt[3]{2})(x-\sqrt[3]{2}w)(x-\sqrt[3]{2}w^2)$
I want to factorize the equation $x^p-q^p$ where $p$ is prime into products of $p$ equations.
For $p=1$ is trivial, and $p=2$ and $p=3$ are easy
$x^2-q^2 = (x-q)(x+q)$
$x^3-q^3 = (x-q)(x-qw)(x-qw^2) $
where $w^2+w+1=0$
How about general case?
Are similar things work? For example
$x^p-q^p = (x-q)(x-qw)(x-qw^2)\cdots (x-qw^{p-1})$
where $w^{p-1}+w^{p-2}+\cdots + w+1=0$
This is your assumed identity:
$$x^p-q^p = (x-q)(x-qw)(x-qw^2)\cdots (x-qw^{p-1})$$
Notice that it trivially holds for $q = 0$. But when $q \neq 0$ we solve $x^p - q^p = 0$, and substitute $x = qy$. Then we get $(qy)^p = q^p$, which has solutions $y^p = 1.$
This has $p$ complex solutions, $1, w, w^2, \dots, w^{p-1}$. Thus we know:
$$(y - 1)(y-w)(y-w^2)\cdots(y-w^{p-1}) = 0$$ Undo the substitution $y = \frac{x}{q}$ and multiply each expression by $q$ to get: $$(x - q)(x-wq)(x-w^2q)\cdots(x-w^{p-1}q) = 0$$