Let $\alpha, \beta, \gamma$ denote the zeros of the polynomial $x^{3}-n x-1$, where $n$ is an integer. The discriminant of the polynomial is defined as

$$\Delta=\Delta\left(1, \alpha, \alpha^{2}\right)=(\alpha-\beta)^{2}(\beta-\gamma)^{2}(\gamma-\alpha)^{2}$$

Prove that, if $\Delta$ is square-free, then $1, \alpha, \alpha^{2}$ is an integral basis for $k=\mathbb{Q}(\alpha)$.

By verifying that

$$\alpha(\alpha-\beta)(\alpha-\gamma)=2 n \alpha+3$$

and further that the field norm of the expression on the left is $-\Delta$, or otherwise, show that $\Delta=4 n^{3}-27$. Hence prove that, when $n=1$ and $n=2$, an integral basis for $k$ is $1, \alpha, \alpha^{2}$.