From time to time I ask about units in Cubic fields. I noticed for $\mathbb{Z}[\sqrt[3]{2}]$ I get an analogue of the Pell equation:

$$ \det \left[ \begin{array}{ccc} a & 2c & 2b \\ b & a & 2c \\ c & b & a \end{array} \right] = a^3 + 2b^3 + 4c^3 - 6abc = 1 $$

without citing the Dirichlet unit theorem.

Clearly the answers are $a + b\sqrt[3]{2} + c\sqrt[3]{4} = (1 + \sqrt[3]{2} + \sqrt[3]{4})^n$ with $n \in \mathbb{Z}$ (since we this number is unit we can have negative exponent.

Except, I have no way of ruling out other families of solutions. And is there any constructive way of solving this, i.e. without taking a wild guess?

In the $\text{deg}=2$ case (such as $x^2 - 2y^2 = 1$) there is an answer using the pell eq and the continued fraction $\sqrt{2} = [1;\overline{2}]$. Can we do something analogous here?

There are continued fractions you can do on triples of numbers. I think the first step here is:

$$ (1, \sqrt[3]{2}, \sqrt[3]{4}) \to (1, \sqrt[3]{2}-1, \sqrt[3]{4}-1) \to \dots $$

not sure what the smallest number is here. I wonder if the familiar story from Pell works here? Does this euclidean algorithm repeat? Maybe this does not lead to finding units.