Determining the Equality of Two Field Extensions

Question

Let $F$ be a field of characteristic $0$. Let $F(\alpha)/F$ be a finite extension of degree not divisible by $3$. Is is true that $F(\alpha^3)=F(\alpha)$? If we assume that they are not equal, since $\alpha$ is a root of $g(x)=x^3-\alpha^3\in F(\alpha^3)[x]$, $$[F(\alpha):F(\alpha^3)]\le3$$ If the degree of $F(\alpha)/F(\alpha^3)=3$, we get a contradiction, as then $3$ divides the degree of $F(\alpha)/F$. If $F(\alpha)\ne F(\alpha^3)$, then we conclude that $[F(\alpha):F(\alpha^2)]=2$, and so $g(x)$ must be reducible in $F(\alpha^3)$. Specifically, we have that $g(x)$ has a linear factor in $F(\alpha^3)$.

If it is true that $F(\alpha^3)=F(\alpha)$, we should get a contradiction from this. However, I can find no reason for why this generates a contradiction. Moreover, my argument is not involving the fact that we're in a characteristic $0$ field. On the other hand, I'm failing to find a counter example. Every extension I can think of for which $F(\alpha^3)\subsetneq F(\alpha)$ is such that $3$ divides the order of the extension $F(\alpha)/F$.

Answer 1

One counterexample with $F=\Bbb Q$ is to take $\alpha=e^{2\pi i/3}$, which is a root of $x^2+x+1$ and so $[\Bbb Q(\alpha):\Bbb Q]=2$ (indeed $\Bbb Q(\alpha)=\Bbb Q(\sqrt{-3})$). But $\alpha^3=1$ and so $\Bbb Q(\alpha^3) = \Bbb Q$.