Factor $x^8-x$ in $\Bbb Z[x]$ and in $\Bbb Z_2[x]$

Question

Factor $x^8-x$ in $\Bbb Z[x]$ and in $\Bbb Z_2[x]$

Here what I get is $x^8-x=x(x^7-1)=x(x-1)(1+x+x^2+\cdots+x^6)$ now what next? Help in both the cases in $\Bbb Z[x]$ and in $\Bbb Z_2[x]$

Edit: I think $(1+x+x^2+\cdots+x^6)$ is cyclotomic polynomial for $p=7$ so it is irred over $\Bbb Z$. Now the problem remains for $\Bbb Z_2[x]$

Answer 1

After my edit I finally got the answer it was under my nose the polynomial $1+\cdots +x^6$ does not have zeros at $0$ and $1$ so it can't be factored as a multiple of a one degree polynomial and one other as a whole.

Edit: But as Lubin mentioned in the comment $1+\cdots +x^6=(1+x+x^3)(1+x^2+x^3)$

Answer 2

To answer a question of yours in the comments, here’s how I thought in factoring $(x^7-1)/(x-1)$ over $\Bbb F_2$:

You’re still talking about the six primitive seventh roots of unity here. But what is the smallest field containing $\Bbb F_2$ that also has seventh roots of unity? That is $\Bbb F_8$, the cubic extension of $\Bbb F_2$. So each of those roots must belong to an irreducible cubic polynomial over $\Bbb F_2$. I happened to know that there are only two irreducible $\Bbb F_2$-cubics, namely $x^3+x^2+1$ and $x^3+x+1$. Our sextic had to be the product of these two.