Find the splitting field $\mathbb K$, the isomorphism type of the Galois group Gal$(\mathbb K / \mathbb Q)$, and the action of its generators on $\mathbb K$ for $x^3 + 3$ (from $\mathbb Q [x]$)
Corollary: Let $K$ be a field, $f ∈ K[x]$ and let $L$ be the splitting field for $f$ over $K$. Write $f$ as $ag_1^{s_1} g_2 ^{s_2} ···g_k^{s_k}$ , where $a ∈ K$ and all $g_i ∈ K[x]$ are monic irreducible and pairwise different. (Since all $g_i$ are monic, $a$ is the leading coefficient of $f$). Let $R$ be the set of roots of $f$ in $L$, and $R_i$ be the set of roots of $g_i, i = 1,...,k$. Then
$\{R_i|1 ≤ i ≤ k\}$ is a partition of $R$; and
the orbits of Gal$(L/K)$ on $R$ are precisely the sets $R_1,...,R_k$.
$\mathbb K =\mathbb Q (3^{1/3} \xi , 3^{1/3} \xi^3, 3^{1/3} \xi^5 )$
$G=\text{Gal} (\mathbb K / \mathbb Q)$ acts faithfully of the set of roots of the given polynomial. Since our given polynomial is irreducible over $\mathbb Q$ (doesn't have any roots in $\mathbb Q$), by the corollary, $G$ has one orbit namely $R=\{3^{1/3} \xi , 3^{1/3} \xi^3, 3^{1/3} \xi^5 \}$.
Action of its generators:
$$ \begin{matrix} \hline ~ & 3^{1/3} \xi & 3^{1/3} \xi ^3 & 3^{1/3} \xi ^5\\ \hline \alpha_1 & 3^{1/3} \xi & 3^{1/3} \xi ^3 & 3^{1/3} \xi ^5 \\ \alpha_2 & 3^{1/3} \xi ^3 & 3^{1/3} \xi ^5 & 3^{1/3} \xi \\ \alpha_3 & 3^{1/3} \xi ^5 & 3^{1/3} \xi & 3^{1/3} \xi ^3 \\ \alpha_4 & 3^{1/3} \xi & 3^{1/3} \xi^5 & 3^{1/3} \xi ^3 \\ \alpha_5 & 3^{1/3} \xi ^5 & 3^{1/3} \xi^3 & 3^{1/3} \xi \\ \alpha_6 & 3^{1/3} \xi ^3 & 3^{1/3} \xi & 3^{1/3} \xi ^5 \\ \hline \end{matrix} $$
I don't know if I have done the above right but all the elements seem to have order three. So the isomorphism type must be something to do with $C_3$...
Please can someone say if I have done the above correct and help me with the isomorphism type part.
I'm not sure what you've called $\xi$ above but to avoid confusion, I'm going to take $\zeta$ to be a primitive cube root of unity. Then the roots are $-3^{1/3},-\zeta 3^{1/3}, -\zeta^2 3^{1/3}$, noting that $(-1)^3=-1$.
If you know what the degree of the splitting field is, then you know the size of the Galois group which should limit the possible isomorphism types. Furthermore, since the Galois group acts on $|R|$ elements, it must embed into the symmetric group on $|R|$ elements which limits your options even further.
Now you are right that since the polynomial is irreducible you have an orbit of size $3$ and hence an element of order $3$ (NB this is only true as $3$ is prime). Can you spot another, rather classical Galois automorphism here though?