For quadratic fields $\mathbb{Q}(\sqrt{d})$, with $d$ squarefree, we know that an integral basis is \begin{align*} \begin{cases} 1, \frac{1 + \sqrt{d}}{2} & d = 1 \mod 4, \\ 1, \sqrt{d} & d = 2, 3 \mod 4, \end{cases} \end{align*} while for pure cubic fields $\mathbb{Q}(\sqrt[3]{m})$, where $m = ab^2$ is cubefree and $\gcd(a,b) = 1$, we know that an integral basis is \begin{align*} \begin{cases} 1, \sqrt[3]{m}, \frac{\sqrt[3]{m^2}}{b} & m \neq \pm 1 \mod 9, \\ 1, \sqrt[3]{m}, \frac{b^2 \pm b^2 \sqrt[3]{m} + \sqrt[3]{m^2}}{3b} & m = \pm 1 \mod 9. \end{cases} \end{align*} However, the cases can be unified to a "guaranteed" integral basis for quadratic fields, where one can show that using $\Delta = \text{disc}(\mathbb{Q}(\sqrt{d}))$, \begin{equation*} 1, \frac{\Delta + \sqrt{\Delta}}{2} \end{equation*} is an integral basis for $\mathbb{Q}(\sqrt{d})$ regardless of $d$ mod 4.
Is anything like this known for the pure cubic case? I'd be happy with the basis depending on quantities associated to $\mathbb{Q}(\sqrt[3]{m})$, such as the discriminant (but this won't be enough, since this doesn't determine pure cubic fields), or on $a$ and $b$ in the factorization of $m$, or anything that doesn't require cases on $m$ mod 9.
