Let $K$ be the field $\mathbb{Q}(\theta)$, where $\theta^3 - 9\theta - 6 = 0$. Much of the heavy work has been done for us: we are given that $(2) = PQ^2$ and $(3) = R^3$ are the prime ideal decompositions of the ideals $(2)$ and $(3)$ in $\mathcal{O}_K$ respectively, so that $P = \langle 2, \theta \rangle$, $Q = \langle 2, \theta + 1 \rangle$, and $R = \langle 3, \theta \rangle$. I want to prove that the ideals $P$, $Q$ and $R$ are principal ideals.
I know that I need to show that $P$, $Q$ and $R$ are generated by a single element, say $a \in \mathcal{O}_K$, through multiplication by the elements of $\mathcal{O}_K$. Would taking each ideal in turn, assuming that they are not principal and seeking a contradiction be a valid way to proceed? Any hints for establishing a method to prove this would be appreciated. Thanks!
Compute norms of $\theta$, $\theta+1$, $\theta+2$, and $\theta+3$ and the result will follow from this information and the prime ideal factorizations of $(2)$ and $(3)$.
If you don't know how computing the norm of an algebraic integer $\alpha$ gives you information about the prime ideal factorization of the principal ideal $(\alpha)$ then you should read your algebraic number theory references more carefully.