How would I go about showing that the splitting fields of each of the following polynomials over $\mathbb{Q}$ are generated by just one root of the polynomial? And then how would I use this information to compute the Galois groups over $\mathbb{Q}$? I'm not sure of what is the best way to proceed.
(a) $x^6 - x^3 + 1$
(b) $x^6 + 3x^5 + 6x^4 + 3x^3 + 9x + 9$.
For the first one, I suggest you go ahead and solve the equation! The roots are $\{ \zeta_{18}, \zeta_{18}^5, \zeta_{18}^7, \zeta_{18}^{11}, \zeta_{18}^{13}, \zeta_{18}^{17} \}$, where $\zeta_{18} = e^{2\pi i / 18}$, so the splitting field is simply $\mathbb Q(\zeta_{18})$.
Now use this fact: if $\alpha$ is a root of a polynomial $f(X) \in \mathbb Q[X]$ and $\sigma$ is an automorphism in $Gal(L : \mathbb Q)$ where $L$ is an extension field of $\mathbb Q$, then $\sigma(\alpha)$ is also a root of $f(X)$. How can you use this fact to show that any automorphism in $Gal(\mathbb Q(\zeta_{18}), \mathbb Q)$ must necessarily send $\zeta_{18}$ to another one of the primitive 18th roots of unity?
Finally, to convince yourself that the Galois group is cyclic, I suggest you think about what you know about the multiplicative group of units modulo $18$, and how this may be relevant. [Hint: "primitive roots".]
For the second one, the roots are $\sqrt[3]{2} \zeta_3^a + \zeta_3^b$ with $a = 0,1,2$ and $b=1,2$. So the splitting field is certainly contained in $\mathbb Q(\sqrt[3]{2}, \zeta_3)$. At the moment I can't tell if the splitting field is the whole of $\mathbb Q(\sqrt[3]{2}, \zeta_3)$ or just a subfield, but all will become clear...
Let's examine the Galois group of $\mathbb Q(\sqrt[3]{2}, \zeta_3)$ over $\mathbb Q$. From the "fact" above, you should be able to show that it is generated by $$ \sigma: \ \ \sqrt[3]{2} \mapsto \sqrt[3]{2} \zeta_3, \ \ \zeta_3 \mapsto \zeta_3$$ $$ \tau: \ \ \sqrt[3]{2} \mapsto \sqrt[3]{2} , \ \ \zeta_3 \mapsto \zeta_3^2$$ Can you see how this follows from the "fact"? And what is the group generated by these elements?
Next, ask yourself: is $\sqrt[3]{2} + \zeta_3$ fixed by any non-trivial subgroup of this group? (I think the answer is no.) Can you use this observation, together with the Galois correspondence, to convince yourself that $\mathbb Q(\sqrt[3]{2} + \zeta_3)$ must be the entire field $\mathbb Q(\sqrt[3]{2}, \zeta_3)$, and therefore, the splitting field is $\mathbb Q(\sqrt[3]{2} + \zeta_3) = \mathbb Q(\sqrt[3]{2}, \zeta_3)$?