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?
Noticing that $5=3+2$, we spot a perfect square
$$\sqrt{\sqrt3(3+2\sqrt3\sqrt2+2)}.$$
Hence,
$$\sqrt[4]{27}+\sqrt[4]{12}.$$
$$\begin{align} \sqrt{5\sqrt{3}+6\sqrt{2}} &= \sqrt{5\sqrt{3}+\left(2\sqrt{6}\right)\sqrt{3}} \\ &= \sqrt[4]{3}\cdot\sqrt{5+2\sqrt{6}} \\ &= \sqrt[4]{3}\cdot\sqrt{2+2\sqrt{6} + 3} \\ &= \sqrt[4]{3}\cdot\sqrt{\frac{4+4\sqrt{6}+\left(\sqrt{6}\right)^2}{2}} \\ &= \sqrt[4]{3}\cdot\sqrt{\frac{\left(2+\sqrt{6}\right)^2}{2}} \\ &= \sqrt[4]{3}\cdot\frac{2\sqrt{2}+2\sqrt{3}}{2}\\ &= \sqrt[4]{3}\cdot\left(\sqrt{2} + \sqrt{3}\right). \end{align}$$
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}
