Prove that $(23, \alpha -10, \alpha -3) = \mathbb{Z}[\alpha]$ Let $\mathcal{O}=\mathbb{Z}[\alpha]$ be a number ring, where $\alpha \in \mathbb{C}, \alpha^3=\alpha+1$. I have proved that
$$23\mathcal{O}=(23, \alpha -10)^2(23, \alpha-3)$$
and that $[\mathbb{Q}(\alpha):\mathbb{Q}]=3$. Now I am asked to prove that $(23, \alpha -10, \alpha -3)= \mathcal{O}$. A basis for $\mathcal{O}$ would be $\{1, \alpha, \alpha^2\}$. It would be enough to show that $1\in (23, \alpha -10, \alpha -3)$. However I've made several attempts and I cannot produce $1$ from a combination of $(23, \alpha -10, \alpha -3)$. Can someone help me with this?

Observe
that not only 23 but also $7=\alpha-3-(\alpha-10)=7$ is a element of $\mathcal{O}$. Note also that $23$ and $7$ are  coprime.
Thus there are integers $u,v$ such that $1=23u+7v$ implying that $1\in \mathcal{O}$.
