How does one prove such an equation? The problem occurred to me while I was trying to solve a problem in planimetry using analytic geometry.
for $b$ between $-\frac{1}2$ and $1$ :
$\sqrt{2+\sqrt{3-3b^2}+b} = \sqrt{2-2b}+ \sqrt{2-\sqrt{3-3b^2}+b}$

First, we rearrange the equation into the equivalent equation
$$
\sqrt{2+b+\sqrt{3-3b^2}}-\sqrt{2+b-\sqrt{3-3b^2}} \overset?= \sqrt{2-2b}
$$
which we want to prove. Both sides are positive (keep in mind we have not yet proven the equality), so we can square both sides, and equivalence still holds:
$$
4+2b-2\sqrt{(2+b)^2-(3-3b^2)}
= \left(\sqrt{2+b+\sqrt{3-3b^2}}-\sqrt{2+b-\sqrt{3-3b^2}}\right)^2
\overset?= 2-2b
$$
Ie, $x^2=y^2$ implies $x=y$ if we know $x,y\ge 0$, otherwise the squares could be equal, but the two sides have opposite signs.
Now, simplify the left-hand side,
$$
2-2b = 4+2b-2\sqrt{1+4b+4b^2}= 4+2b-2\sqrt{(2+b)^2-(3-3b^2)} \overset?= 2-2b,
$$
and we see that there actually is equality.
Note that I have used throughout the assumption that the expression inside square roots are non-negative numbers. That's where limitations on $b$ come from.
