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.

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.
