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.