Show that $\sqrt{4 + 2\sqrt{3}} - \sqrt{3}$ is rational.

Note that $4+2\sqrt{3}=(1+\sqrt{3})^2$

Method 1: Consider this denesting algorithm:

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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$$\sqrt{X\pm Y}=\sqrt{\dfrac {X+\sqrt{X^2-Y^2}}2}\pm\sqrt{\dfrac {X-\sqrt{X^2-Y^2}}2}\tag1$$

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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.

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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,$$\sqrt{4+2\sqrt3}-\sqrt3=1$$

Many questions with sum or difference of square roots can be solved with conjugating.

So, if $s = \sqrt{4 + 2\sqrt{3}} - \sqrt{3}$, and $t = \sqrt{4 + 2\sqrt{3}} + \sqrt{3}$,

$\begin{array}\\ st &=(\sqrt{4 + 2\sqrt{3}} - \sqrt{3})(\sqrt{4 + 2\sqrt{3}} + \sqrt{3})\\ &=4 + 2\sqrt{3}-3\\ &=1 + 2\sqrt{3}\\ \end{array} $

Since $t-s = 2\sqrt{3}$, $st = 1+t-s$ or $s(t+1) = 1+t$, and, by magic, we get $s = 1$ (unless $t+1 = 0$ which it does not since $t > 0$).

I am surprised that this worked so well.

I will suppress my need to generalize and submit this as is.