Let $F(\alpha)$ be an algebraic extension of a real field $F$ by a real $\alpha$. We set $p(x)=Irr(\alpha,x,F)$, the minimal polynomial of $\alpha$ over $F$. Prove that if the degree of $p$ is $100$, then $F(\alpha^3)=F(\alpha)$.
I'm not sure where to start, I think considering $F(\alpha)\supset F(\alpha^3)\supset F$ is a good start, but only because my friend gave me that hint.
By multiplicativity of field extension degree, $[F(\alpha):F(\alpha^{3})]$ divides $[F(\alpha):F] = 100$. Note that $\alpha$ is a root of $X^{3}-\alpha^{3} \in F(\alpha^{3})[X]$. If $X^{3}-\alpha^{3}$ were irreducible over $F(\alpha^{3})$, then we would have $[F(\alpha):F(\alpha^{3})] = 3$, a contradiction. It must then be the case that $X^{3}-\alpha^{3}$ has a root over $F(\alpha^{3})$. The roots of $X^{3}-\alpha^{3}$ over $\mathbb{C}$ are $\alpha, \omega\alpha, \omega^{2}\alpha$, where $\omega$ is a primitive complex $3$rd root of unity. Since $F(\alpha) \subset \mathbb{R}$, this means that the only possibility is $\alpha \in F(\alpha^{3})$, i.e. $F(\alpha) = F(\alpha^{3})$, as desired.