.

(a) $x^7-15$ is irreducible by Eisenstein with $p =3$. $(x^7-1)/(x-1)$ is irreducible since if we replace $x$ by $x+1$ then $((x+1)^7-1)/x$ so with Eisenstein with $p =7$, it's irreducible.\\

(b) Clearly, $K = \Bbb Q(\sqrt[7]{15},\eta)$ where $\eta$ is a primitive $7$th root of unity. Note that $[\Bbb Q(\eta):\Bbb Q] = \varphi(7) =6$ and $[\Bbb Q(\sqrt[7]{15}):\Bbb Q] = 7$. Since $7\mid [\Bbb Q(\sqrt[7]{15},\eta):\Bbb Q]$ and $[\Bbb Q(\sqrt[7]{15},\eta):\Bbb Q]= [\Bbb Q(\sqrt[7]{15},\eta):\Bbb Q(\sqrt[7]{15})][\Bbb Q(\eta):\Bbb Q]$ so that $[\Bbb Q(\sqrt[7]{15},\eta):\Bbb Q(\sqrt[7]{15})] = 7$. Now let $L = \Bbb Q(\sqrt[7]{15})$ and $F =\Bbb Q(\eta)$. Then $K = LF$ with $[K:L] = 7$. Observe that we have an exact sequence

$$0\to \operatorname{Gal}(K/F)\to \operatorname{Gal}(K/\Bbb Q)\to \operatorname{Gal}(F/\Bbb Q)\to 0$$

where the last map is given by restriction. This sequence splits as $\operatorname{Gal}(K/L)\simeq \operatorname{Gal}(F/\Bbb Q)$ where the isomorphism is given by restriction. Hence, $\operatorname{Gal}(K/\Bbb Q)\simeq \operatorname{Gal}(K/F)\rtimes \operatorname{Gal}(F/\Bbb Q)$.\\

Now as $7$ is prime, $(\Bbb Z/7)^\times\simeq\Bbb Z/6$. Let

$$\sigma = \begin{cases} \sqrt[7]{15}\mapsto\sqrt[7]{15}\eta\\

\eta\mapsto\eta\\

\end{cases}, \tau = \begin{cases} \sqrt[7]{15}\mapsto \sqrt[7]{15}\\

\eta\mapsto\eta^c\\

\end{cases}$$

where $c$ is a generator of $\Bbb Z/6$. This gives an action of $\operatorname{Gal}(K/\Bbb Q)$ on $K$.

혹시나 볼 동무들에게 : 증명 생략이 좀 심함.