How to simplify this:$\sqrt{5\sqrt{3}+6\sqrt{2}}$? How to simplyfy this:$\sqrt{5\sqrt{3}+6\sqrt{2}}$.
I know I should use nested radicals formula but which one is $A$ and $B$.Using the fact $A>B^2$ you can find $A$ and $B$.
But $C^2=A-B^2$ isn't a rational number then we have again a nested radical.
What to do?
 A: $$\begin{align}
\sqrt{5\sqrt{3}+6\sqrt{2}}
&= \sqrt{5\sqrt{3}+\left(2\sqrt{6}\right)\sqrt{3}} \\
&= \sqrt[4]{3}\cdot\sqrt{5+2\sqrt{6}} \\
&= \sqrt[4]{3}\cdot\sqrt{2+2\sqrt{6} + 3} \\
&= \sqrt[4]{3}\cdot\sqrt{\frac{4+4\sqrt{6}+\left(\sqrt{6}\right)^2}{2}} \\
&= \sqrt[4]{3}\cdot\sqrt{\frac{\left(2+\sqrt{6}\right)^2}{2}} \\
&= \sqrt[4]{3}\cdot\frac{2\sqrt{2}+2\sqrt{3}}{2}\\
&= \sqrt[4]{3}\cdot\left(\sqrt{2} + \sqrt{3}\right).
\end{align}$$
A: Noticing that $5=3+2$, we spot a perfect square
$$\sqrt{\sqrt3(3+2\sqrt3\sqrt2+2)}.$$

Hence,
$$\sqrt[4]{27}+\sqrt[4]{12}.$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
\color{#f00}{\root{5\root{3} + 6\root{2}}} =
\bracks{\pars{5\root{3} + 6\root{2}}^{2}}^{1/4} =
\color{#f00}{\pars{147 + 60\root{6}}^{1/4}}
\end{align}
