Here is a question I encountered in school:
Show that $\mathbb{Q}\left(\zeta_{21}\right)$ has exactly three subfields of degree 6 over $\mathbb{Q}$. Show that one of them is $\mathbb{Q}\left(\zeta_7\right)$, one is real, and the other is a cyclic extension $K / \mathbb{Q}\left(\zeta_3\right)$. Use a suitable Lagrange resolvent to find $a \in \mathbb{Q}\left(\zeta_3\right)$ such that $K=\mathbb{Q}\left(\zeta_3, \sqrt[3]{a}\right)$.
For the first two parts, I know that $\mathbb{Q}(\zeta_{21})/\mathbb{Q}$ has Galois group $\mathbb{Z}_{21}^\times$, then I can identify order 2 subgroups for the required subfields, where $K = \mathbb{Q}(\zeta+\zeta^{13})$.
I am not so familiar with Lagrange resolvent and need help to see how it is applied here. Thank you!