Let $K=\mathbb{Q}(\sqrt[3]{2})$ I want to show that $U(K)=\{\pm (\sqrt[3]{2} -1)^n : n \in \mathbb{Z}\}$
I know that an integral basis for $K$ is $1, \sqrt[3]{2}, \sqrt[3]{2}^2$
So far I have got: Since the minimal polynomial of $\sqrt[3]{2}$ has one real root and 2 complex roots then K has one real embedding and one complex embedding, so the signature of K is (1,1). The rank of K=1+1-1=1. Then by using Dirichlet's Unit Theorem there is some unit $u$ such that every other unit can be written in the form $\pm u^n$. Therefore $U(K)=\{\pm u^n\}$ but I don't know how to proceed from here.