Factoring $x^6+x^5+x^4+x^3+x^2+x+1$ into two cubics On page 587 of Dummit and Foote, the authors demonstrate producing irreducible polynomials over $\mathbb{F}_p$ recursively. This is after Proposition 18 which states "The polynomial $x^{p^n}-x$ is precisely the product of all the distinct irreducible polynomials in $\mathbb{F}_p[x]$ of degree $d$ where $d$ runs through all the divisors of $n$."
So, to find the irreducible cubics over $\mathbb{F}_2$, we consider the divisors of $\frac{x^8-x}{x(x-1)}=x^6+x^5+x^4+x^3+x^2+x+1$. 
This apparently factors into the two cubics $x^3+x+1$ and $x^3+x^2+1$, but I am having trouble seeing why this is so. How do I factor the polynomial above as such? 
I'd really like to know how to go about factoring the polynomial $x^6+x^5+x^4+x^3+x^2+x+1$ assuming I don't know it factors as the two cubics above. 
 A: A polynomial $p(x)$ of degree $2$ or $3$ is irreducible if and only if it does not have linear factors.  Therefore, it suffices to show that $p(0) = p(1) = 1$.  This quickly tells us that us that $x^3 + x^2 + 1$ and $x^3 + x + 1$ are the only irreducible polynomials of degree $3$ over $\mathbb{F}_2$. So we just try to multiply two of them and compare the result with $x^6+\cdots +x+1$.
A: We want to find $(x^3+ax^2+bx+c)(x^3+dx^2+ex+f)$. We have little choice for $c$ and $f$, as they must be $1$. Also, changing $x$ into $1/x$ and multiplying by $x^6$ should leave the thing fixed: since we get
$$
(x^3+bx^2+ax+1)(x^3+ex^2+dx+1)
$$
we need either $a=b$ and $d=e$ or $a=e$ and $b=d$. This is because of unique factorization, of course.
The first case is easily dismissed:
$$
(x^3+ax^2+ax+1)(x^3+dx^2+dx+1)
$$
would give, for the degree $5$ term, $a+d=1$, so, without loss of generality, $a=0$; however $x^3+1$ is not irreducible.
We remain with
$$
(x^3+ax^2+bx+1)(x^3+bx^2+ax+1)
$$
and we easily get $a=1$ and $b=0$ (or conversely).
