Radical extension with root of cubic polynomial If I take $f(x)$ is an irreducible cubic  over $\mathbb{Q}$ with a root $\alpha$ in a splitting field and given that $\mathbb{Q}(\alpha)$ is a radical extension is it true that $\mathbb{Q}(\alpha) = \mathbb{Q}(b^{1/3})$ for some $b \in \mathbb{Q}$?
We have that $\alpha^q \in \mathbb{Q}$ for some $q$. I.e. $\alpha$ satisfies an equation $g(x) = x^q - b$ with $\alpha^q = b$. As $f$ is the minimal polynomial of $\alpha$ we must have $f$ divides $g$.
Now if it's the case that $q = 3$ then we'd have $f=g$ but that's clearly not the case in general. I'm not quite sure where I am going wrong here - or perhaps the original statement is false.
Another slight point of confusion is that it seems that $g$ is irreducible for any $b$ because it's splitting field is $\mathbb{Q}(\omega, b^{1/q})$ where $\omega$ is a primitive qth root of unity - that extension is Galois and acts transitively on the roots of $g$? But if that were the case then again we'd have $f=g$?
Thanks for any help clearing up this confusion!
 A: The original statement is true. 
We have that if $f$ is an irreducible cubic, and then $\mathbb{Q}(\alpha): \mathbb{Q}$ is radical. Then there exist, from the definition, $\beta_1, \ldots, \beta_n$ with 
$$
\mathbb{Q}(\alpha) = \mathbb{Q}(\beta_1, \ldots , \beta_n),
$$
and for each $i$ we may assume there is some prime $p_i$ such that 
$$
\beta_i^{p_i} \in \mathbb{Q}(\beta_1, \ldots, \beta_{i-1}).
$$
Then by the Tower Law  applied repeatedly to $[\mathbb{Q}(\beta_1,\ldots, \beta_n):\mathbb{Q}]$ we get
$$
[\mathbb{Q}(\beta_1,\ldots, \beta_n):\mathbb{Q}]$ = \prod_{i=1}^n[\mathbb{Q}(\beta_1,\ldots, \beta_i):\mathbb{Q}(\beta_1,\ldots, \beta_{i-1})] = \prod_{i=1}^n p_i.
$$
But then as $[\mathbb{Q}(\beta_1,\ldots, \beta_n):\mathbb{Q}] = [\mathbb{Q}(\alpha):\mathbb{Q}] = 3$, we see that, by prime factorisation, there is only one element $\beta$ with $\beta^3 \in \mathbb{Q}$ and 
$$
\mathbb{Q}(\alpha) = \mathbb{Q}(\beta).
$$
But the important point is that this does not imply that $\alpha = \beta$, but instead that there are rational numbers $q_0, q_1, q_2$ with 
$$
\alpha = q_0 + q_1 \beta + q_2 \beta^2.
$$
EDIT:
Because $\mathbb{Q}(\alpha)$ is a radical extension of degree 3, by the tower law analysis above we know that there is some $\beta$ which generates $\mathbb{Q}(\alpha)$. 
Then the minimal polynomial of $\beta$ divides $x^3 - \beta^3$. But since $[\mathbb{Q}(\alpha):\mathbb{Q}] = 3$, then $[\mathbb{Q}(\alpha):\mathbb{Q}] = 3$, so the minimal polynomial of $\beta$ must be of degree 3 and so equals $x^3 -\beta^3$.
