"Let $L_{1} = \mathbb{Q}(\alpha)$, where $\alpha^3-18\alpha-6=0$
and $L_{2} = \mathbb{Q}(\alpha)$, where $\alpha^3-36\alpha-78=0$
(i) Prove $L_{1}$ and $L_{2}$ are cubic fields
(ii) Prove that $1, \alpha, \alpha^2$ is an integral basis of $O_{K_1}$, resp. $O_{K_2}$, so $O_{K_1} = \mathbb{Z}[\alpha]$, $O_{K_2} = \mathbb{Z}[\alpha]$"
So part (i) is pretty simple, by showing that the polynomials $X^3-18X-6$ (for $L_{1}$) and $X^3-36X-78$ (for $L_{2}$) are irreducible over $\mathbb{Q}$, then we know the degree of the field extension is $3$ in both cases, so both are cubic fields. To prove this, I used Eisenstein's Criterion with primes $2$ and $3$
For part (ii), what I have is:
We know $1, \alpha, \alpha^2$ form a $\mathbb{Q}$-basis of $L_{1}$ and $L_{2}$. I've considered the discriminant of both polynomials, getting $22356=2^{2}.3^{5}.23$ and $10692=2^{2}.3^{5}.11$. Even though these are not squarefree, as 2 and 3 fulfil the Eisenstein's Criterion for both polynomials, we only have to consider $23$ and $11$, which are squarefree, so $O_{K_1} = \mathbb{Z}[\alpha]$, $O_{K_2} = \mathbb{Z}[\alpha]$ and so $1, \alpha, \alpha^2$ is an integral basis of $O_{K_1}$ and $O_{K_2}$
Am I missing anything? I'm most confused by what it means by '$1, \alpha, \alpha^2$ is an integral basis of $O_{K_1}$, resp. $O_{K_2}$'