If $$2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{a}+\sqrt{b}$$, where $$a$$ and $$b$$ are natural numbers, then the value of $$a+b$$ is

Since $$\sqrt{13+\sqrt{48}}=\sqrt{13+4\sqrt{3}}=2\sqrt{3}+1$$, the next radical is $$\sqrt{4-2\sqrt{3}}=\sqrt{3}-1$$. Thus the outer expression becomes $$2\sqrt{2+\sqrt{3}}$$. Since $$\sqrt{2+\sqrt{3}}=\frac{\sqrt{6}+\sqrt{2}}{2}$$, the expression equals $$\sqrt{6}+\sqrt{2}$$. Hence $$a+b=6+2=8$$.