Let $\alpha = \sqrt[3]{2}$, which is integral, and $K = \mathbb{Q}(\alpha)$ be a number field of degree $3$, with basis ${\{1, \alpha, \alpha^2}\}$ and let $\mathcal{O}_K$ denote its ring of integers.
Let $\theta = \alpha - 2$, which is also integral, then I have shown that there exists no $\tau \in \mathcal{O}_K$ such that $\tau = \frac{x + y\theta + z\theta^2}{3}$, where $x,y,z \in {\{0,1,2}\}$, with $\tau \neq 0 $.
I now want to show that there exists no such $\tau' \in \mathcal{O}_K$ such that $\tau' = \frac{x' + y'\alpha + z'\alpha^2}{3}$, where $x', y', z' \in {\{0,1,2}\}$, with $\tau' \neq 0$.
I've been trying to do this for a while and not got anywhere.
Firstly, I know that $\mathbb{Z}[\alpha] = \mathbb{Z}[\theta]$, which I think could prove useful.
My way of attacking this was to assume that there does exist $\tau' \in \mathcal{O}_K$ of the required form, and tried to show that this implies that there is a $\tau \in \mathcal{O}_K$ of the required form, which would lead to a contradiction.
Hence, assume there exists a $\tau' = \frac{x' + y'\alpha + z'\alpha^2}{3}$, where $x', y', z' \in {\{0,1,2}\}$ in the ring of integers, not all $x',y',z' = 0$. I then tried to fiddle around with ring operations to show that $\tau$ must then be in the ring of integers, but to no avail.
Could anyone give me a hint or any help? Thank you.
Assume $\tau' = \frac{x'+y'\alpha + z' \alpha^2}{3} \in \mathcal{O}_K$. If we prove that $\tau' = 0$ then taking the contrapositive of this statement means $\tau' \neq 0 \implies \tau' \not\in \mathcal{O}_K$.
We have $\alpha = \theta +2$ and so, as Connor Harris suggested in the comments, we can rewrite this as $\tau' = \frac{(x'+2y'+4z') + (y'+2z')\theta + z'\theta^2}{3}$. Since $\tau' \in \mathcal{O}_K$ by your previous results this forces the coefficients of $\frac{1}{3},\ \frac{\theta}{3},$ and $\frac{\theta^2}{3}$ to be zero. Hence we get the equations \begin{array}{rcl} x'+2y'+4z' & = & 0 \\ y'+2z' & = & 0 \\ z' & = & 0 \end{array}
Therefore we must have $x' = y' = z' = 0$, and so $\tau' = 0$, as required.