Show that $x^6+x^3+1$ is irreducible over $\mathbb{F}_2$ This is part of a larger question where I'm supposed to determine if $x^6+x^3+1$ is irreducible over $\mathbb{F}_2$, $\mathbb{F}_3$, $\mathbb{F}_{19}$ and  $\mathbb{Q}$.
For $\mathbb{F}_3$ and $\mathbb{F}_{19}$ it's easy enough to just find roots and factorise, while over $\mathbb{Q}$ I can substitute and then apply Eisenstein.
I'm struggling to think of how to show it's irreducible over $\mathbb{F}_2$(I'm pretty sure it is). Clearly it doesn't have any roots in $\mathbb{F}_2$ , and so all that's left to show is that no polynomials of degree 2 or 3 divide it, but I have no idea how to go about doing that.
Any help you could offer would be really appreciated.

Let's look for factors of degree $3$. Suppose (over $\mathbb F_2$)
$$x^6 + x^3 + 1 = (x^3 + a_2 x^2 + a_1 x + a_0)(x^3 + b_2 x^2 + b_1 x + b_0)$$
Since $a_0 b_0 = 1$, we must have $a_0 = b_0 = 1$.
By looking at the $x^5$ and $x^1$ terms, you must have $a_2=b_2$ and $a_1=b_1$.
 But then the $x^3$ term on the right is $0$.
The case of a factor of degree $2$ is similar.
