Is it true that all units in $\mathbb{Z}[\sqrt[3]{2}]$ are of form $\pm(1+\sqrt[3]{2}+\sqrt[3]{4})^n$ for some integer $n$ ?
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I can't read that. Is that a three? An s? – RougeSegwayUser Aug 12 '14 at 21:08
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Rings of Gauss ? – Maman Aug 12 '14 at 21:12
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Yes, this is true. The unit group has rank $1$ by Dirichlet's theorem, and the group $\mu_K$ is $\{ \pm 1\}$. We have that $u=1+\sqrt[3]{2}+\sqrt[3]{4}$ is a fundamental unit. Hence the unit group is given by $\pm u^{\mathbb{Z}}$. A detailed proof can be found in the notes of Keith Conrad, Theorem 2.

Dietrich Burde
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