<s>[INST] How do I show that $\sqrt{5+\sqrt{24}} = \sqrt{3}+\sqrt{2}$ According to wolfram alpha this is true: $\sqrt{5+\sqrt{24}} = \sqrt{3}+\sqrt{2}$
But how do you show this? I know of no rules that works with addition inside square roots.
I noticed I could do this:
$\sqrt{24} = 2\sqrt{3}\sqrt{2}$
But I still don't see how I should show this since $\sqrt{5+2\sqrt{3}\sqrt{2}} = \sqrt{3}+\sqrt{2}$ still contains that addition
 [/INST] On can easily discover the denesting using my simple radical denesting algorithm.
$\ w = 5+\sqrt{24}\,$ has norm $\,n = ww' = 5^2-24 = 1.\,$ Subtracting out $\,\sqrt{n}=1\,$ yields $\,4+\sqrt{24}.$ 
This has trace $\,t = 8,\,$ so dividing $\,\sqrt{t} = 2\sqrt{2}\,$ out of $\,4+\sqrt{24}=4+2\sqrt{6}\,$ yields
$$ \frac{4+2\sqrt{6}}{2\sqrt{2}}\,=\, \frac{2+\sqrt{6}}{\sqrt{2}\ } \,=\, \sqrt{2}+\sqrt{3}$$
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