I want to find the fundamental units of $K = \mathbb{Q}(\sqrt[3]{12})$. The extension degree is $n = 3$, we have $r_1 = 1$ real embedding and $2r_2 = 2$ complex embeddings. So by Dirichlet's theorem, $\mathbb{Z}_K^\times \cong \mu_K \oplus \mathbb{Z}u$ where $\mathbb{Z}_K$ is $K$'s ring of integers and $\mu_K$ is the group of roots of unity in $K$. I first proved that $B = \{1, \theta, \frac{\theta^2}{2}\}$ where $\theta = \sqrt[3]{12}$.
$\pm1$ are the only rational roots of unity. To find other roots of unity in $K$ let $\zeta_n$ be a complex primitive $n$-th root of unity in $K$. Then $\mathbb{Q} \subseteq \mathbb{Q}(\zeta_n) \subseteq \mathbb{Q}(\sqrt[3]{12})$. Then $[\mathbb{Q}(\sqrt[3]{12}): \mathbb{Q}] = 3 = [\mathbb{Q}(\sqrt[3]{12}): \mathbb{Q}(\zeta_n)]\times[\mathbb{Q}(\zeta_n):\mathbb{Q}]$. But 3 is prime and $[\mathbb{Q}(\zeta_n): \mathbb{Q}] = \phi(n)$ is even when $n \geq 2$. So $\mu_K = \{\pm1\} \cong \mathbb{Z}_2$.
Now I want to find fundamental unit $u$. This is where I'm stuck. I know $v$ is a unit $\iff$ $N_{K/\mathbb{Q}}(v) = \pm1$. So I took an element $v = a + b\theta + c\frac{\theta^2}{2}$ and calculated its norm. This gave me the following equation:
$$a^3 + 18c^3 + 12b^3 - 18abc = 1$$
I didn't follow a very formal method to find solutions for $a, b, c$. I just substituted things that satisfied the equation. For example, $a^3 = 1$ mod $3$ so I just set $a = 1$. Then $2b^3 = 0$ mod $3$. But since I don't want a real unit, I set $b = 3$ instead of $b = 0$ and so on. This gave me a unit $u = \frac{2 + 6\theta - 3\theta^2}{2}$. I found a paper that stated that $v = \frac{110 + 48\theta + 21\theta^2}{2}$ is a fundamental unit for this field. I calculated $u^{-1}$ using Euclid's algorithm and found that $u^{-1} = v$ so what I've found is a fundamental unit. I just don't know how to prove it.