All intermediate subfields I have this multi-part problem in Galois Theory.

Find all the subfields of the splitting field $L$ of the polynomial $X^3+3 \in \mathbb{Q}[x]$. (Exhibit explicit generators for these).
Prove or disprove: $X^6 + 3$ splits over $L$.

I know that $L$ is Galois extension over $\mathbb{Q}$.
Thanks.
 A: As you commented $[L:\mathbb Q]=6$. This is because the polynomial $X^3+3$ is irreducible (thanks to Eisenstein criterion, $p=3$) so $\sqrt[3]{-3} = \alpha$ has degree $3$ over $\mathbb Q$.
The splitting field $L$ is $\mathbb Q(\alpha, \omega)$ where $\omega$ is the third primitive root of unity. Since $\omega$ has degree $2$ (it's minimal polynomial is $x^2+x+1$), then $[L:\mathbb Q]=6$ (because $2$ and $3$ are coprime).
Also the extension $L/\mathbb Q$ is normal and separable, hence it is a Galois extension.
Define the following automorphisms of $L$:
\begin{equation}
\rho:L\rightarrow L, \begin{cases}
\rho(\alpha) = \alpha\omega\\
\rho(\omega) = \omega
\end{cases}
\qquad 
\sigma:L\rightarrow L, \begin{cases}
\sigma(\alpha) = \alpha\\
\sigma(\omega) = \omega^2
\end{cases}
\end{equation}
They are well defined automorphisms of $\text{Gal}(L/\mathbb Q))$. Now $\rho$ has order $3$, $\sigma$ has order $2$, $\langle \rho\rangle \cap \langle \sigma \rangle = \emptyset$ and $\sigma\rho\sigma^{-1} = \rho^{-1}$. This easily implies (cardinality) that:
$$
\text{Gal}(L/\mathbb Q)) = \langle \rho, \sigma \ | \ \rho^3=\sigma^2=id, \sigma\rho\sigma^{-1} = \rho^{-1}\rangle \cong S_3
$$
Thanks to the Galois corrispondance theorem we have that the intermediate subfields are in bijection with the subgroups of $\text{Gal}(L/\mathbb Q)=S_3$.
We have only $6$ subgroups, so we have only $6$ intermediate subfields that are:
\begin{gather}
S_3 \quad \longleftrightarrow  L^{S_3} = \mathbb Q\\
\langle \rho \rangle \quad \longleftrightarrow L^{\langle \rho \rangle} = \mathbb Q(\omega)\\
\langle \sigma \rangle \quad \longleftrightarrow  L^{\langle \sigma \rangle} = \mathbb Q(\alpha)\\
\langle \sigma\rho \rangle \quad \longleftrightarrow  L^{\langle \sigma\rho \rangle} = \mathbb Q(\alpha\omega)\\
\langle \sigma\rho^2 \rangle \quad \longleftrightarrow  L^{\langle \sigma\rho^2 \rangle} = \mathbb Q(\alpha\omega^2)\\
\{id\} \quad \longleftrightarrow  L^{\{id\}} = L\\
\end{gather}

Let's denote with $K$ the splitting field of $X^6+3$ over $\mathbb Q$. $K = \mathbb Q(\sqrt[6]{-3}, \omega_6) = \mathbb (\sqrt[6]{-3},\omega)$ where $\omega_6$ is the sixth primitive root of unity. 
Observe that $\left(\sqrt[6]{-3}\right)^2 = \sqrt[3]{-3}$; hence $L\subset K$.
Now if we show that the degree $[K:\mathbb Q]=6$, then we have $L=K$ (a containment and equal degree). 
Observe that $\left(\sqrt[6]{-3}\right)^3 = i\sqrt{3}$, and $\mathbb Q(\omega) = \mathbb Q(i\sqrt 3)$. Hence $K=\mathbb Q(\sqrt[6]{-3},\omega) = \mathbb Q(\sqrt[6]{-3})$ and this field has degree $6$ over $\mathbb Q$ because $X^6+3$ is irreducible by Eisenstein.
So $L=K$ and $X^6+3$ splits over $L$.
