Show Galois group of $f(t) = t^5 - 3$ over $\mathbb Q$ is solvable I am asked to show that the galois group of $f(t) = t^5 -3 $ over $\mathbb Q$ is solvable, and it is heavily implied that I should be making use of the derived series. 
$f$ is irreducible with discriminant not a square, so $Gal(f) \in \{H_{20}, S_5\}$. As $S_5$ is not solvable, it is implied then that $Gal(f) = H_{20}$, however I don't know how to conclusively show this. 
Additionally, given that $Gal(f) = H_{20}$ how might I actually go about computing the derived series from here? I am aware that the derived subgroup is containing in any normal subgroup of $Gal(f)$, but I am not aware of what the normal subgroups of $H_{20}$, and trying to find one by hand has been difficult. 
Is there any slightly easier way to deal with this?
 A: $\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$$\newcommand{\Q}{\mathbb{Q}}$$\newcommand{\Gal}{\mathrm{Gal}}$You could note that the splitting field of $f$ over $\Q$ is $\Q(\alpha, \omega)$, where $\alpha = \sqrt[5]{3}$, and $\omega$ is a primitive $5$-th root of unity. Now $\Size{\Q(\alpha) : \Q} = 5$ and $\Size{\Q(\omega) : \Q} = 4$, and since the two indices are coprime you obtain $\Size{\Q(\alpha, \omega) : \Q} = 20$.
Then you may note that since $\Q(\omega)/\Q$ is a Galois extension, $\Gal(\Q(\alpha, \omega)/\Q(\omega))$ is a normal subgroup of $\Gal(\Q(\alpha, \omega)/\Q)$ of order $5$ and index $4$.
A: In general, the Galois group of $x^n-a$ is explicitly known, see here:
Computing the Galois group of polynomials $x^n-a \in \mathbb{Q}[x]$
Proposition [Jacobson, Velez]: One has $G\cong \mathbb{Z}/n \rtimes (\mathbb{Z}/n)^{\times}$ if and only if $n$ is odd, or $n$ is even and $\sqrt{a}\not\in \mathbb{Q}(\zeta_n)$. 
Such Galois groups are solvable, being a semidirect product of solvable groups. Here we have the case $(n,a)=(5,3)$.
A: Here is another take.
It suffices to prove that Galois group has order $20=2^2 \cdot 5$ because of Burnside theorem:

Every group of order $p^a q^b$ is solvable.

