I am studying for my Galois theory final for tomorrow (and I'm really getting burned out), I need help with the following question:
Galois group $G$ of $f(x)=(x^3-2)(x^5-1)$ over $\mathbb{Q}$.
Let $\alpha = \sqrt[3]{2}$ and let $\omega_3$ be the primitive cubed root of unity, and $\omega_5$ the primitive $5$th root of unity. The splitting field for $f$ is $E=\mathbb{Q}(\alpha,\omega_3, \omega_5)$. We know that $E:\mathbb{Q}$ is separable since $\text{char}\mathbb{Q}=0$ and that it is normal since it is a splitting field. Hence $E:\mathbb{Q}$ is Galois. So $|G|=|E:\mathbb{Q}|$. It is easy for me to show that $\text{Gal}((x^3-2)/\mathbb{Q}) \cong S_3$ and that $\text{Gal}((x^5-1)/\mathbb{Q}) \cong C_4$.
But how can I determine $\text{Gal}(f)$? How can I find the size of the extension $|E:\mathbb{Q}|$? I know it is divisible by $6$ and $4$.
Edit: I see people mentioning the result about cartesian products. I have not seen this result. Given that this is a past exam question I would be interested in a proof that does not use the result
The Galois group of $f=gh$ is equal to the cartesian product of the Galois group of $g$ and the Galois group of $h$ iff the splitting field of $g$ over $\mathbb{Q}$ is disjoint with the splitting field of $h$ over $\mathbb{Q}$. I.e. you have to determine whether $\mathbb{Q}(\omega_3,\sqrt[3]{2})\cap\mathbb{Q}(\omega_5)=\mathbb{Q}$. If shown, you have that Gal$(\mathbb{Q}(\omega_3,\sqrt[3]{2},\omega_5)/\mathbb{Q})\cong S_3\times C_4$.