Splitting field of $f=x^6+x^5+1$, orders of elements, and minimal polynomials. 
Let $F$ be the splitting field over $\Bbb{Z}_2$ of $f=x^6+x^5+1$, an irreducible polynomial over $\Bbb{Z}_2$.
We know that:
$\bullet$ The Galois group of $F$ over $\Bbb{Z}_2$ is cyclic and is generated by the Frobenius autormorphism.
$\bullet$ The lattice of the subgroups of $Gal(F/\Bbb{Z}_2)$ is isomorphic to that of $\Bbb{Z}_6$. Let's denote the one of degree 2 over $\Bbb{Z}_2$ as $\Bbb{Z}_2(\omega)$ and the one of degree 3 over $\Bbb{Z}_2$ as $\Bbb{Z}_(\beta)$.
$\bullet$ There are four intermediate fields  with orders $|F|=64$, $|\Bbb{Z}_2(\beta)|=8$, $|\Bbb{Z}_2(\omega)|=4$, and $|\Bbb{Z}_2|=2$ as additive groups.
a) For each intermediate field $E$ (including $F$), what are the possible multiplicative orders of the units $\gamma$ for which $E=\Bbb{Z}_2(\gamma)$?
b) What is the minimal polynomial of an element of order $9$?

a) Since fields are also additive subgroups, the orders of the multiplicative groups is just Euler's totient function. So $|F^{\times}|=\phi(63)=(7-1)(3-1)(3)=36$, $|\Bbb{Z}_2(\beta)|=2^2=4$, $|\Bbb{Z}_2(\omega)|=2$, and $|\Bbb{Z}_2|=2$.
Orders of the units must divide the orders of the multiplicative groups, so for the possibilities, we have (let $F=\Bbb{Z}(\alpha)$)
$|\alpha|=2,3,6,4,9$
$|\beta|=2,4$
$|\omega|=2$
$1$ is not an option, because otherwise we would have $\Bbb{Z}_2(\gamma)=\Bbb{Z}_2$.
b) We know that the minimal polynomial of that element (call it $\epsilon$), must divide $x^9-1$.
We also know that $x^9-1=\Psi_1(x)\Psi_3(x)\Psi_9(x)$, where $\Psi_k(x)$ is the $k$th cyclotomic polynomial.
Now we need to find $\Psi_9(x)$ in order to use this.
$x^9-1=(x-1)(x^2+x+1)\Psi_9(x)$
$\implies \Psi_9(x) = \frac{x^9-1}{(x-1)(x^2+x+1)}$
$\implies \Psi_9(x) = \frac{(x-1)(x^8+x^7-x^6+x^5-x^4+x^3-x^2+x-1)}{(x-1)(x^2+x+1)}$
$\implies \Psi_9(x) = \frac{x^8+x^7-x^6+x^5-x^4+x^3-x^2+x-1}{x^2+x+1}$.
However, when I tried to do long division, I kept getting a remainder of -2...and I'm not sure why. Doesn't $x^8+x^7-x^6+x^5-x^4+x^3-x^2+x-1$ for sure need to be divisible by $x^2+x+1$? Because we know for sure that $x^9-1$ equals $\Psi_1(x)\Psi_3(x)\Psi_9(x)$, right?
Thank you in advance

$(x^9-1)/(x-1)$ is not $x^8+x^7-x^6+x^5-x^4+x^3-x^2+x-1$, but $x^8+x^7+\cdots +x+1$.
Indeed, 
$$
x^9-1=(x^6+x^3+1)(x^2+x+1)(x-1),\quad \frac{x^9-1}{x-1}=(x^6+x^3+1)(x^2+x+1).
$$
