Let $\mathbb{Q} \subset K$ denote the splitting field of $P = (X^2-5)(X^3-2)$.
I have to find $\alpha \in \mathbb {C} $ such that $K = \mathbb{Q}(\alpha)$. We know that such an $\alpha$ exists by the primitive element theorem.
The roots of $P$ in $\mathbb {C}$ are $\sqrt5, -\sqrt5, 2^{1/3}, j2^{1/3}$ and $j^{2}2^{1/3}$ where $j = e^{2i\pi/3}$.
So we can already say that $K = \mathbb{Q}(j, \sqrt5, 2^{1/3})$.
I've already shown that $\mathbb{Q}(\sqrt5, 2^{1/3})= \mathbb{Q}(\sqrt5 + 2^{1/3})$ so I know that $K = \mathbb{Q}(j, \sqrt5 + 2^{1/3})$. How do I find a generator of K now ?
In the proof of the primitive element theorem, I saw that $K = \mathbb{Q}(t.j + \sqrt5 + 2^{1/3})$ for all but finitely many $t \in \mathbb {Q}$, so there are many possibilities.
Let $\beta=\sqrt{5}+2^{\frac{1}{3}}$, claim $\alpha=j+\beta$. In fact $\mathbb{Q}[j+\beta]$ is an extension of degree 2 over $\mathbb{Q}[\beta]$ since this last extension is real and does not contain $j$. You had already proved that the real extension has degree 6 over the field of rational numbers. We conclude that the bigger extension has degree 12