Legal of nested radicals

Question

$\sqrt{20+ \sqrt{96}+\sqrt{12}} = \sqrt9 + \sqrt6 + \sqrt3 - \sqrt2$. Hi, What legal in math cause this statement?

Answer 1

Use the equations \begin{eqnarray*} \sqrt{\alpha}+\sqrt{\beta}=\sqrt{\alpha+\beta+2\sqrt{\alpha\beta}} \\ \sqrt{\alpha}-\sqrt{\beta}=\sqrt{\alpha+\beta-2\sqrt{\alpha\beta} } \end{eqnarray*} In your case ... \begin{eqnarray*} \sqrt{9}+\sqrt{6}+\sqrt{3}-\sqrt{2} & =&\sqrt{15+6\sqrt{6}}+\sqrt{5-2\sqrt{6}} \\ & =&\sqrt{20+4 \sqrt{6}+2 \sqrt{(15+6\sqrt{6})(5-2\sqrt{6})}} \\ & =& \sqrt{20+\sqrt{96}+\sqrt{12}} \end{eqnarray*}