Discriminant of $\Bbb Q(\sqrt[3]{2})$

I'm not sure that this is what your teacher had in mind, but it allows a quick calculation.

Let $f \in \mathbb{Q}[x]$ monic of degree $n$ be the minimum polynomial of $\alpha$ and let $K=\mathbb{Q}(\alpha)$.

1. The discriminant of $\mathbb{Z}[\alpha]$ is $(-1)^{\frac{1}{2}n(n-1)}N^K_{\mathbb{Q}}(f'(\alpha))$.
2. The field norm $N^K_{\mathbb{Q}}$ is multiplicative.
3. $N^K_{\mathbb{Q}}(b) = b^n$ for $b \in \mathbb{Q}$.
4. $N^K_{\mathbb{Q}}(\alpha)$ is $(-1)^n$ times the constant term of $f$.

Putting these together for $f = x^3 - 2$ (using $\mathcal{O}_K = \mathbb{Z}[\alpha]$ here, which is nontrivial) yields

The absolute value of the discriminant of $\mathbb{Q}(\sqrt[3]{2})$ is $N(3\alpha^2) = N(3)N(\alpha)^2 = 3^3 \cdot 2^2$.

Concerning the necessary background, the article The splitting field of $x^3-2$ by K. Conrad does explain this in detail, see Theorem $2$ on page $4$. Taking the determinant of the matrix $(\sigma_i(x_j))^2_{i,j}$ gives the discriminant, where $\{x_1,x_2,\ldots, x_n\}$ is an integral basis of the number field $K$ over $\mathbb{Q}$, and the $\sigma_i$ the embeddings.Here $n=3$. Another method is explained here; and using the trace matrix for the discriminant of a cubic number field has been computed here.