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? 
 A: Here's a hint to give you the kind of idea involved:
The innermost term $3-2\sqrt2$ can be rewritten as $1^2-2\cdot1\cdot\sqrt 2+(-\sqrt{2})^2=(1-\sqrt 2)^2$ which cancels the square root it lies inside. Now simply and apply the same strategy of recognizing squares to keep canceling square roots until you are left with a simple expression.
Edit: How do we recognize these squares?
The core idea lies in the "mixed term". Our goal is to make the term inside the current innermost radical look like either $(a+b)^2=a^2+b^2+2ab$  or $(a-b^2)=a^2+b^2-2ab$ which have middle terms $2ab$ or $-2ab$ respectively. So in our example, we had $-2\sqrt 2$, so you know that we're going to get something of the form $(a-b)^2$ with $$ab=\sqrt 2.$$ However, we need $$a^2+b^2=3.$$
At this point you one of the $a$ or $b$ as $\sqrt 2$ since it shows prominently in the $ab$ term. So $a^2+(\sqrt 2)^2=3$, so $a$ needs to be $1$. This checks out with $ab=\sqrt 2$. This last bit, you have to eyeball it a little bit. 
A: Observe that
$$(\sqrt{2}-1)^2=2-2\sqrt{2}+1=3-2\sqrt{2}\qquad\implies\qquad\color{blue}{\sqrt{3-2\sqrt{2}}=\sqrt{2}-1}$$
Then,
$$\color{blue}{\sqrt{10+4\sqrt{3-2\sqrt{2}}}}=\sqrt{10+4(\sqrt{2}-1)}=\sqrt{6+4\sqrt{2}}=\sqrt{4+4\sqrt{2}+2}=\color{blue}{\sqrt{(2+\sqrt{2})^2}}$$
So
\begin{align}
\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}&=\sqrt{23-6(2+\sqrt{2})}\\
&=\sqrt{11-6\sqrt{2}}\\
&=\sqrt{9-6\sqrt{2}+2}\\
&=\sqrt{(3-\sqrt{2})^2}\\
&=3-\sqrt{2}
\end{align}
Finally
\begin{align}
\sqrt{9-2\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}}&=\sqrt{9-2(3-\sqrt{2})}\\
&=\sqrt{3+2\sqrt{2}}\\
&=\sqrt{2}+1
\end{align}
A: Suppose that the term $\sqrt{3-2\sqrt2}$ can be written as $\sqrt a + b$ for some $a,b\in\mathbb Z$. We can equate these expressions then square both sides:
$$\sqrt{3-2\sqrt2} = \sqrt a + b\\
3 - 2\sqrt2 = a+b^2 + 2b\sqrt a$$
and from here you can see that the solution is $(a,b) = (2,-1)$. This gives you a methodical approach that you can apply recursively.
A: 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.  
