How to show $\sqrt{4+2\sqrt{3}}-\sqrt{3} = 1$ I start with $x=\sqrt{4+2\sqrt{3}}-\sqrt{3}$, then
$\begin{align*}
x +\sqrt{3} &= \sqrt{4+2\sqrt{3}}\\
(x +\sqrt{3})^2 &= (\sqrt{4+2\sqrt{3}})^2\\
x^2 + (2\sqrt{3})x + 3 &= 4+ 2\sqrt{3}\\
x^2 + (2\sqrt{3})x - 1 - 2\sqrt{3} &= 0
\end{align*}$
So I have shown that there is some polynomial whose solution is $x=\sqrt{4+2\sqrt{3}}-\sqrt{3}$, but I have not shown it to be 1.

Here’s an approach that doesn’t require you to spot a not-terribly-obvious factorization. Multiplying by the conjugate to get rid of some of the square roots is a fairly natural thing to do, and
$$\left(\sqrt{4+2\sqrt3}-\sqrt3\right)\left(\sqrt{4+2\sqrt3}+\sqrt3\right)=1+2\sqrt3\;.$$
Thus, the desired result holds if and only if 
$$1+2\sqrt3=\sqrt{4+2\sqrt3}+\sqrt3\;,$$
or
$$1+\sqrt3=\sqrt{4+2\sqrt3}\;,$$
which is easily verified.
