Find a unit in $\mathbb{Q}(\sqrt[3]{6})$ and show that this field has class number $h=1$.
I am done with the first part which is relatively simple:
Suppose that $\varepsilon$ is a unit in $\mathbb{Q}(\sqrt[3]{6})$. Then we have $\varepsilon=c+b\sqrt[3]{6}+a\sqrt[3]{6^2}$, since the integral base of $\mathbb{Q}(\sqrt[3]{6})$ can be written as $\{1,\sqrt[3]{6},\sqrt[3]{6^2}\}$. Thus, $\varepsilon=c+b\sqrt[3]{6}+a\sqrt[3]{6^2},$ $\sqrt[3]{6}\varepsilon=6a+c\sqrt[3]{6}+b\sqrt[3]{6^2},$ $\sqrt[3]{6^2}\varepsilon=6b+6a\sqrt[3]{6}+c\sqrt[3]{6^2}.$
As we see it, the system of equations with variable $\varepsilon$ has only zero solution, since $\{1,\sqrt[3]{6},\sqrt[3]{6^2}\}$ is a base. Then $\det\left( \begin{array}{ccc} c-\varepsilon & b & a \\ 6a & c-\varepsilon & b \\ 6b & 6a & c-\varepsilon \\ \end{array} \right) $ is the minimal polynomial of $\varepsilon$. Since $\varepsilon$ is a unit in $\mathbb{Q}(\sqrt[3]{6})$ if and only if $N(\varepsilon)=\pm1$, we take $\varepsilon=0$ in the above polynomial, and $\det\left( \begin{array}{ccc} c & b & a \\ 6a & c & b \\ 6b & 6a & c \\ \end{array} \right)=\pm1. $ Compute the determinant we find that $a=33,~b=60,~c=109$ is one of the solutions. Hence a unit in $\mathbb{Q}(\sqrt[3]{6})$ is of the form $\varepsilon=109+60\sqrt[3]{6}+33\sqrt[3]{6^2}$.
For the second part of the problem, I have no idea how to show that $\mathbb{Q}(\sqrt[3]{6})$ is a principal ideal domain.
Any comment will be appreciated!