Multiple Nested Radicals $\sqrt{9-2\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}}$ 
I have no idea how to unnest radicals, can anyone help? 

Work from the inside out.
Let $\sqrt{3 - 2\sqrt{2}} = \sqrt{a} - \sqrt{b}$, where $a$ and $b$ are rational numbers.  Squaring both sides yields
$$3 - 2\sqrt{2} = a + b - 2\sqrt{ab}$$
Then 
\begin{align*}
a + b & = 3 \tag{1}\\
-2\sqrt{ab} & = -2\sqrt{2} \tag{2}
\end{align*}
Dividing equation by $-2$ yields
$$\sqrt{ab} = \sqrt{2}$$
Squaring both sides of the equation yields
$$ab = 2$$
Hence,
$$b = \frac{2}{a}$$
Substituting for $b$ in equation 1 yields
\begin{align*}
a + \frac{2}{a} & = 3\\
a^2 + 2 & = 3a\\
a^2 - 3a + 2 & = 0\\
(a - 1)(a - 2) & = 0
\end{align*}
Hence, $a = 1$ or $a = 2$.  If $a = 1$, then $b = 2$.  However, $\sqrt{1} - \sqrt{2} = 1 - \sqrt{2} < 0$, but $\sqrt{3 - 2\sqrt{2}} > 0$.  Thus, $a = 2$ and $b = 1$.  Hence,
$$\sqrt{3 - 2\sqrt{2}} = \sqrt{2} - \sqrt{1} = \sqrt{2} - 1$$
Substituting $\sqrt{2} - 1$ for $\sqrt{3 - 2\sqrt{2}}$ yields
$$\sqrt{10 + 4\sqrt{3 - 2\sqrt{2}}} = \sqrt{10 + 4(\sqrt{2} - 1)} = \sqrt{6 + 4\sqrt{2}}$$ 
Let $\sqrt{10 + 4\sqrt{2}} = \sqrt{c} + \sqrt{d}$, where $c$ and $d$ are rational numbers.  Continue.  
