Galois group of $x^3 - 2 $ over $\mathbb Q$

A brute-force way to see it is easy enough. The roots of $X^3 - 2$ are $\sqrt[3]{2},\; \omega\sqrt[3]{2},\; \omega^2\sqrt[3]{2}$, so the splitting field of your polynomial is $K = \mathbb{Q}(\sqrt[3]{2},\omega)$. You are asking why it's legitimate to send $\sqrt[3]{2}$ to $\omega\sqrt[3]{2}$. Well, let's try it. Let $\sigma: K \to K$ be a map that sends $\sqrt[3]{2}$ to $\omega\sqrt[3]{2}$. How are we going to extend this to an element of the Galois group? We need $\sigma$ to be an automorphism that fixes $\mathbb{Q}$.

Our field $K$ is a $\mathbb{Q}$-vector space generated by six basis elements: $1,\; \sqrt[3]{2},\; \sqrt[3]{4},\; \omega,\; \omega\sqrt[3]{2},\; \omega\sqrt[3]{4}$. If we can define how $\sigma$ acts on those, we can extend linearly to the whole space: that is, we can extend $\sigma$ such that $\sigma(a+b) = \sigma(a) + \sigma(b)$ for all $a$ and $b$. We also know that, if we define $\sigma$ sensibly, we should get that $\sigma(ab) = \sigma(a) \sigma(b)$. By "sensibly" here, I mean that $\sigma$ should send each element of $K$ to one of its conjugates - that is, it should send $x$ to any $y$ that satisfies the same minimal polynomial as $x$. We also know that it's enough to make sure $\sigma$ is multiplicative on the basis, and in fact we only need to define it on $\sqrt[3]{2}$ and $\omega$, since the rest will then follow automatically by multiplicativity.

So we have a few options from what we've deduced so far. $\sigma$ can send $\sqrt[3]{2}$ to any of $\sqrt[3]{2},\; \omega\sqrt[3]{2},\; \omega^2\sqrt[3]{2}$, and it can send $\omega$ to either of $\omega$ and $\omega^2$. So let's say $\sigma(\sqrt[3]{2}) = \omega\sqrt[3]{2}$ (as we wanted) and $\sigma(\omega) = \omega$ (no good reason for this choice, but we had to make one). Where do all the other basis elements go? Can you convince yourself now that $\sigma$ is an element of the Galois group?

(What would have happened if we'd chosen $\sigma(\omega) = \omega^2$?)

What you're probably missing is that, although the real cube root of 2 stands out, all three complex cube roots of 2 are algebraically the same, that is, they cannot be told apart, and so can be permuted at will. Hence $S_3$.

Contrast this with the seemingly analogous situation of, say, the 6-th roots of unity. Here the roots are not the same algebraically, because $x^6-1$ is not irreducible.

I don't know what kind of argument you'd like to see, but the standard way to compute the Galois group of an irreducible cubic (characteristic not $2$) is to look at the sign of its discriminant $\Delta = \prod_{i \neq j} (r_i - r_j)$. It has a square root $\prod_{i < j} (r_i - r_j)$ in a splitting field of the polynomial which is multiplied by $1$ when an even permutation is applied to the roots and multiplied by $-1$ when an odd permutation is applied to the roots.

By the fundamental theorem of Galois theory, it follows that the Galois group of an irreducible polynomial of degree $n$ is contained in $A_n$ if and only if the discriminant is a square. In your example, the discriminant of the cubic $x^3 + px + q$ is given by $-4p^3 - 27q^2 = -108$ which is not a square, so the Galois group is not contained in $A_3 \cong C_3$, hence must be $S_3$.

Another way is to use Dedekind's theorem that if an irreducible polynomial $f$ factors into distinct irreducible factors $f = \prod f_i \bmod p$, then the Galois group of $f$ has a permutation with cycle type $(\deg f_1, \deg f_2, ...)$. Thus to show that there is a $3$-cycle in the Galois group of $x^3 - 2$ it suffices to find a prime with respect to which this polynomial is irreducible, and this is easy since for cubics irreducibility is equivalent to not having a root. $p = 7$ works.