Is $\mathbb{Q}(\zeta + 1/\zeta)/\mathbb{Q}$ a radical extension?

Question

I am trying to answer part (d). I know the minimum polynomial of $\alpha$ is:

$f(x)=x^3+x^2-2x-1$ which has roots: $\alpha, \alpha^2-2$ and $-\alpha^2-\alpha+1$.

so $[\mathbb{Q}(\alpha):\mathbb{Q}]=3$

so if this extension is radical then some power of $\alpha$ is in $\mathbb{Q}$ as there are no intermediate fields as their dimension would have to divide 3.

Now I know that $f(x)$ is soluble which, I think, implies that $\mathbb{Q}(\alpha)$ is soluble and so is contained in a radical field.

Also if $\alpha^n = a \in \mathbb{Q}, n\ge 1$ then $f(x)\mid x^n-a$

I have tried $\alpha^7$ as this seems like an obvious choice but it is hard to simply a massive polynomial and it's not clear its in $\mathbb{Q}$ but from here I'm stumped.

Answer 1

If $K/\Bbb Q$ were radical, there would be some $a\in\Bbb Q$ such that $\sqrt[3]a$ is irrational and would belong to $K$(why?). As $K/\Bbb Q$ is Galois and $K\leq\Bbb R$ this gives a contradiction.