= Solution
There is an error in the printed request: the cubic is not irreducible at $n=2$, and three elements cannot be a field <basis> in that case. The integral-basis assertion is valid under the necessary hypothesis $[\mathbb Q(\alpha):\mathbb Q]=3$.
Under that hypothesis, $\alpha$ is an <algebraic integer> and $\mathbb Z[\alpha]\subseteq\mathcal O_k$ has finite index $m$. Change of <basis> in the <trace pairing> gives the discriminant-index identity
$$
\Delta(1,\alpha,\alpha^2)=m^2\operatorname{disc}(\mathcal O_k).
$$
The <field discriminant> is an <integer>. If the left side is square-free, $m^2$ cannot divide it unless $m=1$. Thus \b[$1,\alpha,\alpha^2$ is an <integral basis> when the cubic is irreducible and its <discriminant> is square-free].
For any root, $\alpha^3=n\alpha+1$ gives
$$
\alpha(\alpha-\beta)(\alpha-\gamma)=\alpha f'(\alpha)=3\alpha^3-n\alpha=2n\alpha+3.
$$
Since $\alpha+\beta+\gamma=0$, $\alpha\beta+\beta\gamma+\gamma\alpha=-n$ and $\alpha\beta\gamma=1$,
$$
\prod_{r\in\{\alpha,\beta,\gamma\}}(2nr+3)=27-12n^3+8n^3=27-4n^3.
$$
On the other hand, the product of $r f'(r)$ is $-\Delta$: the product of the roots is one, and each of the three unordered root pairs contributes a negative squared difference. Therefore, without needing irreducibility for this polynomial calculation,
$$
\boxed{\Delta=4n^3-27.}
$$
The three-root product is the <field norm> only when the cubic is the minimal polynomial.
At $n=1$, the polynomial has no rational root (the only candidates are $\pm1$), so it is irreducible, and its <discriminant> is $-23$. Hence \b[$1,\alpha,\alpha^2$ is an <integral basis>].
At $n=2$,
$$
x^3-2x-1=(x+1)(x^2-x-1).
$$
If $\alpha=-1$, then $k=\mathbb Q$ with <integral basis> $\{1\}$. If $\alpha=(1\pm\sqrt5)/2$, then $k=\mathbb Q(\sqrt5)$ and \b[$\{1,\alpha\}$ is an <integral basis>], since this quadratic order has square-free <discriminant> five; also $\alpha^2=\alpha+1$. In neither case is the three-element list a <basis>. Its <discriminant> five is instead that of the degree-three algebra $\mathbb Q[x]/(x^3-2x-1)$; a <basis> there does not repair the printed assertion about the field $k$.
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