Compute the Unit and Class Number of a pure cubic field $\mathbb{Q}(\sqrt[3]{6})$ 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!

The first idea for computing units in such fields is finding a generator of a purely ramified prime. Here $2 - \sqrt[3]{6}$ has norm $2$, hence
$$ (2 - \sqrt[3]{6})^3 = 2(1 - 6\sqrt[3]{6} + 3\sqrt[3]{6}^2) $$
is $2$ times a unit. 
Finding an element generating the prime ideal above $3$ is more difficult, but it turns out that $\beta = 3 + 2\sqrt[3]{6} + \sqrt[3]{6}^2$ is such an element with norm $3$. As above you now get
$$ (3 + 2\sqrt[3]{6} + \sqrt[3]{6}^2)^3 
   = 3(109+60\sqrt[3]{6} +33\sqrt[3]{6}^2), $$ 
and you get the unit you mentioned in your question. Actually we have
$$ \frac1{1 - 6\sqrt[3]{6} + 3\sqrt[3]{6}^2} = 
        109+60\sqrt[3]{6} +33\sqrt[3]{6}^2 . $$
Finding elements of norms $5$ and $7$ is rather easy, which then shows that the ring ${\mathbb Z}[\sqrt[3]{6}]$ is principal.
