abstract algebra - Efficient way to find $[mathbb{Q}(sqrt{5}, sqrt{3}, sqrt{2}): mathbb{Q}(sqrt{3}, sqrt{2})]$

I want to show rigorously that this is 2. I'm sure there's a faster way than by trying to see if $\sqrt{5} = a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6}$ (linear combination of the basis elements for $\mathbb{Q}(\sqrt{3}, \sqrt{2})$). I'm just not sure what.