Find the Galois group of $x^3-5$ over $\mathbb{Q}$. In this case, the roots of $x^3-5$ are $\{\sqrt[3]{5},\omega\sqrt[3]{5},\omega^2\sqrt[3]{5}\}.$ I think $\mathbb{Q}(\sqrt[3]{5},\omega\sqrt[3]{5})$ is the splitting field of $x^3-5.$ Then, $[\mathbb{Q}(\sqrt[3]{5},\omega\sqrt[3]{5}):\mathbb{Q}]=3,$ I think the automorphisms are:
$$id$$
$$\sigma:\sigma(\sqrt[3]{5})=-\sqrt[3]{5}$$
$$\tau:\tau(\omega\sqrt[3]{5})=-\omega\sqrt[3]{5}$$
I'm not sure about what I have done. But how to determine the Galois group based on this? And how to determine the solvability of this polynomial?
 A: We have the root $\sqrt[3] 5$. Now, factor $x^3-5$ using that root:
$$x^3-5=x^3-\sqrt[3] 5 x^2+\sqrt[3] 5 x^2-\sqrt[3] 25 x+\sqrt[3] {25} x-5=(x-\sqrt[3] 5)(x^2+\sqrt[3] 5 x+\sqrt[3] {25})$$
Now, the other two roots must be from this quadratic and using the quadratic formula, we can see they are not in $\Bbb{Q}(\sqrt[3] 5)$. Thus, to split $x^3-5$ in $\Bbb{Q}$, we need a root of a cubic ($\sqrt[3] 5$ from $x^3-5$) and a root of a quadratic (whatever the root of $x^2+\sqrt[3] 5 x+\sqrt[3]{25}$ is). Thus, the degree of the splitting field is $3*2=6$ and this is also the order of the Galois group.
Since this is a cubic, the Galois group is a subgroup of $S_3$. Now, what subgroup of $S_3$ has an order of $6$? The answer to that question is the Galois group.
A: $x^3-5$ is irreducible, so the Galois group is a transitive subgroup of $S_3$.  The only possibilities are $C_3$ and $S_3$.
But your automorphism $\tau$ has order $2$, while $C_3$ has no elements of order $2$.  So the Galois group is $S_3$.

We can also make a similar argument, using the Galois correspondence, by observing that $\mathbb{Q}[\omega]$ is contained in the splitting field, so there is a subgroup of index $2$.
Note that your claim $[\mathbb{Q}(\sqrt[3]{5},\omega\sqrt[3]{5}):\mathbb{Q}]=3$ is false, in fact $[\mathbb{Q}(\sqrt[3]{5},\omega\sqrt[3]{5}):\mathbb{Q}]=6$, so the Galois group has order $6$.
