<s>[INST] Show that $\sqrt{4 + 2\sqrt{3}} - \sqrt{3}$ is rational. 
Show that $\sqrt{4 + 2\sqrt{3}} - \sqrt{3}$ is rational.

I've tried to attempt algebra on this problem. I noticed that there is some kind of nesting effect when trying to solve this. Please help me to understand how to attempt to denest this number.
Any help would be greatly appreciated.
 [/INST] Method 1: Consider this denesting algorithm:

Denested square roots: Given a radical of the form $\sqrt{X\pm {Y}}$ with $X,Y\in\mathbb{R}$ and $X>Y$, we have a possible simplification as$$\sqrt{X\pm Y}=\sqrt{\dfrac {X+\sqrt{X^2-Y^2}}2}\pm\sqrt{\dfrac {X-\sqrt{X^2-Y^2}}2}\tag1$$

Using $(1)$ on $\sqrt{4+2\sqrt3}$, we have$$\sqrt{4+2\sqrt3}=\sqrt{\dfrac {4+2}2}+\sqrt{\dfrac {4-2}2}=\sqrt3+1\tag2$$
So the original expression becomes$$\color{brown}{\sqrt{4+2\sqrt3}}-\sqrt3=\color{brown}{\sqrt3+1}-\sqrt3=1$$
Which is rational.

Method 2: If you don't like $(1)$ and think it's too complicated, I present you an alternative method. Simply set $\sqrt{4+2\sqrt{3}}-\sqrt3$ equal to a variable and simplify!
Here, we have$$\begin{align*} & \sqrt{4+2\sqrt3}-\sqrt3=\alpha\\ & \sqrt{4+2\sqrt3}=\alpha+\sqrt3\\ & 4+2\sqrt3=(\alpha+\sqrt3)^2\\ & 4+2\sqrt3=\alpha^2+3+2\alpha\sqrt3\end{align*}$$
To solve for $\alpha$, we have $2\alpha\sqrt3=2\sqrt3\implies\alpha=1$. Checking with the other half, $\alpha^2+3=1+3=4$ holds. Hence,$$\sqrt{4+2\sqrt3}-\sqrt3=1$$
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