How can I prove algebraically that $\sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}$? I found out that $\sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}$, by reverse engineering:
$$\sqrt{6}-\sqrt{2} =
 \sqrt{ (\sqrt{6}-\sqrt{2} )^{2} } =  \sqrt{ (\sqrt{6}-\sqrt{2} ) (\sqrt{6}-\sqrt{2} )}= \sqrt{ 6-\sqrt{6×2}-\sqrt{6×2}+2}=\sqrt{ 8-2\sqrt{12}}=\sqrt{ 8-4\sqrt{3}}$$
But how can I do it the other way, also for other similar problems?
I tried multiple problem solving sites, but none could simplify $\sqrt{8-4\sqrt{3}}$ further then $2\sqrt{2-\sqrt{3}}$

If you didn't know the answer already for $\sqrt{8-4\sqrt{3}}$, you could start with $$8-4\sqrt{3}=(\sqrt{a}-\sqrt{b})^2$$
$$\implies8-4\sqrt{3}=a-2\sqrt{ab}+b$$
$$\implies a+b=8$$ and $$\sqrt{ab}=2\sqrt{3}\implies ab=12$$
and solve for $a$ and $b$
