Consider the pure cubic field $K=\mathbb{Q}(\sqrt[3]{10})$ then as $10\equiv 1 \pmod 9$ then the integral basis for $K$ is of the form $\{1,\sqrt[3]{10},\frac{1+\sqrt[3]{10}+\sqrt[3]{10^2}}{3}\}$. And the ring of integers
$\mathcal{O}_{K}=\mathbb{Z}\bigoplus \mathbb{Z} \sqrt[3]{10} \bigoplus \mathbb{Z}\frac{1+\sqrt[3]{10}+\sqrt[3]{10^2}}{3}$ which is the maximal order in $K$.
I also have found out that the field $K$ has a power integral basis that is of the form
$\biggl\{1,\frac{1+\sqrt[3]{10}+\sqrt[3]{10^2}}{3},\bigl(\frac{1+\sqrt[3]{10}+\sqrt[3]{10^2}}{3}\bigr)^2\biggr\}$ which gives the ring
$R=\mathbb{Z}\bigoplus \mathbb{Z} \frac{1+\sqrt[3]{10}+\sqrt[3]{10^2}}{3} \bigoplus \mathbb{Z}\Bigl(\frac{1+\sqrt[3]{10}+\sqrt[3]{10^2}}{3}\Bigr)^2$
My teacher says that the above rings are the same thing and they represent the same unique maximal order in $K$ (if I understood him correctly)
To show that these are the same I computed the index $[\mathcal{O}_{K}:R]=1$ which means that $\mathcal{O}_{K}=R$ but the maximal order is unique for every number field, can I conclude then that $\mathcal{O}_{K}$ and $R$ are the same rings (maximal order), but expressed by different integral bases?
Thank you