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 $2\sqrt{2-\sqrt{3}}$

In the general case let solve: $$\sqrt{x\pm y}=\sqrt{a}\pm\sqrt{b}$$
Since LHS exsits and is positive we will also assume $x\ge y$ and $a\ge b$.
By squaring expressions we get the system $\begin{cases}x+y=a+b+2\sqrt{ab}\\x-y=a+b-2\sqrt{ab}\end{cases}\iff \begin{cases}x=a+b\\y=2\sqrt{ab}\end{cases}$
Solving for $a,b$ we get $\begin{cases}a=\frac 12 x+\frac 12\sqrt{x^2-y^2}\\b=\frac 12 x-\frac 12\sqrt{x^2-y^2}\end{cases}$
And there you have it:

$$\sqrt{\vphantom{|}x\pm y}=\sqrt{\frac 12 x+\frac 12\sqrt{x^2-y^2}}\pm\sqrt{\frac 12 x-\frac 12\sqrt{x^2-y^2}}$$

Applying for $x=8$ anf $y=4\sqrt{3}$ we get $\sqrt{x^2-y^2}=\sqrt{64-16\times 3}=\sqrt{16}=4$
So $\sqrt{8-4\sqrt{3}}=\sqrt{4+2}-\sqrt{4-2}=\sqrt{6}-\sqrt{2}$.
