$\mathbb{Q}(\sqrt{6},\sqrt{10},\sqrt{15})=\mathbb{Q}(\alpha, \beta),$ for some $ \alpha $ and $\beta$?

Question

Let $\mathbb{Q}\subseteq\mathbb{Q}(\sqrt{6},\sqrt{10},\sqrt{15})$ be a extension of $\mathbb{Q}$, i need to show the elements of $Gal(\mathbb{Q}(\sqrt{6},\sqrt{10},\sqrt{15})\mid\mathbb{Q})$ the Galois group asociate to this extension, i know that is enogth knowing were the root of the minimal polynimials of $\sqrt{6},\sqrt{10}$ and $\sqrt{15}$ are asigned, to create the automorphisms, but i had notice that $\sqrt{6}=\sqrt{3}\sqrt{2},\sqrt{10}=\sqrt{5}\sqrt{2}$ and $\sqrt{15}=\sqrt{3}\sqrt{5}$, my question is, there is a way to show that the field $\mathbb{Q}(\sqrt{6},\sqrt{10},\sqrt{15})$ is of the form $\mathbb{Q}(\alpha,\beta)$ for some $\alpha,\beta$ depending of $\sqrt{2}, \sqrt{3}$ and $\sqrt{5}$, any clue?

Answer 1

To answer the question in the title, $\mathbb{Q}(\sqrt{6},\sqrt{10},\sqrt{15})=\mathbb{Q}(\sqrt{6},\sqrt{10})$ because $\sqrt{6}\sqrt{10}=2\sqrt{15}$ implies $\sqrt{15} \in \mathbb{Q}(\sqrt{6},\sqrt{10})$.