= Solution
The polynomial $T^3+3T+3$ is Eisenstein at $3$, so part (ii) gives
$$
3\nmid[\mathcal O_K:\mathbb Z[\alpha]].
$$
For a cubic $T^3+aT+b$, the polynomial discriminant is $-4a^3-27b^2$. Here
$$
\operatorname{disc}(1,\alpha,\alpha^2)
=-4\cdot3^3-27\cdot3^2
=-351=-3^3\cdot13.
$$
The <discriminant-index formula for an integral lattice> says
$$
-351=[\mathcal O_K:\mathbb Z[\alpha]]^2d_K.
$$
Any prime dividing the index must therefore occur to at least the second power in $351$; the only possibility is $3$, which has already been excluded. The index is one, and hence
$$
\boxed{\mathcal O_K=\mathbb Z[\alpha]}.
$$
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