Let $K=\mathbb{Q}(\theta)$ where $\theta$ is a root of $X^3-2X-2$. Integral basis for $\mathcal{O}_{K}$? 
Let $K=\mathbb{Q}(\theta)$ where $\theta$ is a root of $X^{3}-2X-2$. Compute an integral basis for $\mathcal{O}_{K}$.

I have computed the discriminant as $\Delta_{K}=-2^2.19$. I want to apply the algorithm in the book of Frazer Jarvis Algebraic Number Theory. We know that if $\beta=\frac{u_0+u_1\theta+u_2\theta^2}{2}$ is an integral element then its trace and norm is an integer. $Trace(\beta)=\frac{3u_0+4u_2}{2}$. Since $0 \leq u_i<2$ for any i, $u_0=0.$ Then the only possible $\beta$ are : $\frac{\theta}{2},\frac{\theta^2}{2},\frac{\theta+\theta^2}{2}$. But when I computed the norm of all possiblities, Norms are 1/4, 1/2, 1/4 respectively, then they are not integer.
Is there any computational mistakes or what is wrong?

The reason why neither of the norms is an integer is because the elements listed are not algebraic integers. Indeed, for example $\frac{\theta + \theta^2}2$ has $4X^3 - 8X^2 - 4X - 1$ as a minimal polynomial. In fact, it turns out that the ring of integers $\mathcal O_K$ is equal to $\mathbb Z[\theta]$, which can be deduced by eliminating all possibilites that come from applying the algorithm. So an integral basis of it is $\{1,\theta,\theta^2\}$.
There is another way to compute the disciminant $d_K$ of the extension $K/\mathbb Q$, which is much more theory dependent, but doesn't involve long computations. In shortly, the only unknown prime power of $d_K$ is the power of $2$. To find out, you make use of the fact that $X^3 - 2X - 2$ is an Eisenstein polynomial with respect to $2$, compute the $2$-adic different of $K/\mathbb Q$ and deduce that $4$ is the highest power of $2$ dividing $d_K$. So as $d_K = \text{disc} f$ we deduce that $\mathcal O_K = \mathbb Z[\theta]$.
However, I suspect that this method might be too advanced. But, if you want me to, I can include the computation steps, as well as giving you references for the theoretic stuffs.
