= Solution
Let $\theta=\sqrt[3]3$. The power-<basis> <discriminant> is
$$
\operatorname{disc}(1,\theta,\theta^2)=\operatorname{disc}(x^3-3)=-27\cdot9=-243.
$$
If $\mathbb Z[\theta]$ were a proper sublattice of $\mathcal O_K$, part (a) would produce a prime $p$ whose square divides $243$ and a new integral element with denominator $p$. The only possible prime is $3$. But $x^3-3$ is Eisenstein at $3$, and part (b) says that every integral element of $(1/3)\mathbb Z[\theta]$ already belongs to $\mathbb Z[\theta]$. This is a contradiction. Hence
$$
\boxed{\mathcal O_K=\mathbb Z[\sqrt[3]3],\qquad
(1,\sqrt[3]3,\sqrt[3]9)\text{ is an integral basis}.}
$$
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