Multiple Nested Radicals

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.

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}

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.