Field extensions of degree 10 over the rationals.

[Theorem.]

Homework Statement

Show that if $\alpha$ is real and has degree 10 over $\mathbb{Q}$ then
$\mathbb{Q}[\alpha]=\mathbb{Q}[\alpha^3]$

Homework Equations

The Attempt at a Solution

It is clear that $\mathbb{Q}[\alpha^3]\subset \mathbb{Q}[\alpha]$. This gives us the
sequence of fields $\mathbb{Q}\subset \mathbb{Q}[\alpha^3]\subset \mathbb{Q}[\alpha]$. Since these are finite extensions, we then have $[\mathbb{Q}[\alpha]:\mathbb{Q}]=10=[\mathbb{Q}[\alpha]:\mathbb{Q}[\alpha^3]][\mathbb{Q}[\alpha^3]:\mathbb{Q}]$.
Since $x^3-\alpha^3\in \mathbb{Q}[\alpha^3][x]$ is of degree 3 and has $\alpha$ as a root, $[\mathbb{Q}[\alpha]:\mathbb{Q}[\alpha^3]]\leq 3$. Since it must also divide 10, it must be 1 or 2. The goal then is to show that it cannot be 2 (or then that $[\mathbb{Q}[\alpha^3]:\mathbb{Q}]$ cannot be 5.)
I have tried going further on this point, but I think I'm just getting stubborn and missing something subtle. Any hints would be appreciated.
-Theorem

 

[Theorem.]

Okay I just got an idea. We can suppose towards a contradiction that $\alpha \notin \mathbb{Q}[\alpha^3]$ Then the polynomial $x^3-\alpha^3$ is irreducible in $\mathbb{Q}[\alpha^3]$ since
$\alpha$ is real by assumption and so $\mathbb{Q}[\alpha^3]$ is contained in $\mathbb{R}$ and the other two roots of the polynomial are non real (and thus it cannot be factored over $\mathbb{Q}[\alpha^3]$. Buth then this leads to a contradiction since then this forces $[\mathbb{Q}[\alpha]:\mathbb{Q}[\alpha^3]]=3$. which is impossible.