How to compute the Galois Group of $(x^3-3)(x^4-2)$ over $\mathbb Q$ So I know the splitting field is $\mathbb Q(\sqrt[4]2, i ,\sqrt[3]3,\zeta_3 )$ where $\zeta_3=-\frac12+\frac{\sqrt3}{2}i$ is the root of unity. And the order of the Galois Group equals the degree of the field extension, which is $48$ in this case. But this is quite a large group so I am lost on where to go for the next step. Is there some standard way to compute the Galois group like this?
 A: Here is a computation of the Galois group, you can look at the emphasized steps without reading the whole solution if you want to compute it by yourself :)
Throughout I set $\alpha = \sqrt[4]{2}$, $\beta = \sqrt[3]{3}$ and I use the notation $j = \zeta_3$. With these notations the roots of the polynomial are $\lbrace \alpha , -\alpha , i\alpha , -i\alpha , \beta , j\beta , j^2 \beta \rbrace$.
Step 1: Split the big group. As suggested by leoli1, you can show that this big Galois group $G$ is actually isomorphic to the product $G_1\times G_2$, where $G_1:= Gal (x^3-3)$ and $G_2 = Gal(x^4-2)$.
An elementary way of seeing this is by looking at the relations between the roots. If $\sigma$ is an element of $G$ seen as a permutation of the roots, then you can check that $\sigma (\alpha ) \in \lbrace \beta , j\beta ,j^2\beta \rbrace$ cannot be. For example if $\sigma (\alpha ) =\beta$ then $\sigma (-\alpha ) = -\beta$ so $-\beta$ must be a root of the polynomial: contradiction.
Checking all the possibilities you will eventually get that "the $\alpha$'s are sent to a root in $\alpha$, and same for the $\beta$'s" (this is informal).
You can also argue that the big extension $\mathbb{Q} (\alpha ,i,\beta , j)$ is actually the composite $\mathbb{Q} (\alpha ,i) \mathbb{Q} (\beta , j)$ and that these are linearly disjoint (their intersection is $\mathbb{Q}$). Then a general result tells you that $G \simeq G_1\times G_2$.
Step 2: Computing $G_2$. Start with $G_2$, which is the Galois group of $x^3-3$. This is a sugroup of $S_3$ (symmetric group) that has order $[\mathbb{Q} (\beta , j) : \mathbb{Q} ] = 6$. But this is exactly the order of $S_3$, so $G_2\simeq S_3$.
Step 3: Computing $G_1$. This one is a little bit trickier. We know that $G_1$ is a subgroup of $S_4$ of order $[\mathbb{Q} (\alpha , i) : \mathbb{Q} ] = 8 = 2^3$. But the order of $S_4$ is $4! = 2^3 \cdot 3$. Hence $G_1$ is a $2$-Sylow of $S_4$, and this fully determines $G_1$ up to isomorphism because all the $2$-Sylows are conjugates. For $S_4$ the $2$-Sylows are isomorphic to $D_8$, the diedral group of order $8$ (sometimes denoted $D_4$), so $G_1\simeq D_8$.
However since $8$ is not too big we can compute $G_1$ by hand. Note that if $\sigma\in G_1$, then $( \sigma (\alpha ) , \sigma (i\alpha ))$ determine $\sigma$ as then $\sigma (-\alpha ) = -\sigma (\alpha )$ and similarly for $\sigma (-i\alpha )$. However $\sigma (\alpha )$ does not determine $\sigma (i\alpha )$ since the only relation we have between these two is $(i\alpha)^2 = \alpha^2$. Then if you order the roots as $\lbrace \alpha , -\alpha ,i\alpha ,-i\alpha\rbrace$ you can check by hand all the possible combinations and this gives you:
$$G_1 = \lbrace id, (1\;2), (3\;4), (1\; 2)(3\;4),(1\;3)(2\; 4),(1\;4)(2\;3), (1\;3\;2\;4)  , (1\;4\;2\;3) \rbrace $$
Then you can check that this group is generated by $(1\;4\;2\;3)$ and $(3\;4)$, and recognize the presentation of $D_8$ with two generators.
Conclusion: $G\simeq D_8\times S_3$.
