Suppose that we have a equation like this:
$$\sqrt{a+b+2\sqrt{ab}}$$ or $$\sqrt{a+b-2\sqrt{ab}}$$
In order to rationalize it, we can apply the formula:
$$\sqrt{a} + \sqrt{b} = \sqrt{a+b+2\sqrt{ab}}$$ or $$\sqrt{a} - \sqrt{b} = \sqrt{a+b -2\sqrt{ab}}$$, where $a>b$
My question is:
Is it possible to rationalize the form like this:
$$\sqrt{10+2(\sqrt{15}+\sqrt{10}+\sqrt{6})}$$ ?
Thanks
Notice that there are $3$ terms with $2$ so probably it will be of form $(a+b+c)^2$ $$(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)\\a^2+b^2+c^2=10\\ab=\sqrt{15}\\ac=\sqrt{10}\\bc=\sqrt{6}\\b^2+c^2=10-a^2\\a^2b^2+a^2c^2=25\\a^2(b^2+c^2)=25\\a^2(10-a^2)=25\\10a^2-a^4-25=0\\a^4-10a^2+25=0\\(a^2-5)^2=0\\a=\sqrt{5}\\ab=\sqrt{15}\implies b=\sqrt{3}\\bc=\sqrt{6}\implies c=\sqrt{2}$$ So the answer is $\sqrt{2}+\sqrt{3}+\sqrt{5}$