= Solution
A rational root of the monic cubic would be one of $\pm1,\pm3$, and substitution excludes all four. A reducible cubic over $\mathbb Q$ would have a rational root, so the <polynomial> is irreducible and
$$
\boxed{[K:\mathbb Q]=3.}
$$
Its <polynomial discriminant> is $-4(-1)^3-27(3)^2=-239$. The integer 239 is prime: it has no prime divisor among $2,3,5,7,11,13$, the primes not exceeding its square root.
The <discriminant-index formula for an integral lattice> for an integral primitive element is
$$
\operatorname{disc}(1,\alpha,\alpha^2)
=[\mathcal O_K:\mathbb Z[\alpha]]^2\operatorname{disc}(K).
$$
Both discriminants are integers, and the left side is $-239$. Its squarefreeness forces the positive index to be one. Since $\alpha$ is an <algebraic integer> by its monic equation,
$$
\boxed{\mathcal O_K=\mathbb Z[\alpha].}
$$
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