<s>[INST] Simplifying expression $\sqrt{6+2\sqrt{5}} - \sqrt{6-2\sqrt{5}}.$ I'm stuck. Squaring it will change it's value. Is there any general method of simplifying expressions of the form $$\sqrt{a+b}-\sqrt{a-b} = c \ \ ?$$
 [/INST] Here is how you de-nest double layer square roots.  All variables below are assumed rational.
Start with
\begin{equation} \tag{1}
\sqrt{a+ \sqrt{b}}
= \sqrt {x} + \sqrt{y},
\end{equation}
which has the conjugate relation
\begin{equation} \tag{2}
\sqrt{a - \sqrt{b}}
= \sqrt{x} -\sqrt{y},
\end{equation}
where $x\ge y $ wlog.
Multiply these two relations together, and use the difference of squares factorization:
\begin{equation} \tag{3}
\sqrt {a^2-b}
\equiv r
= x-y.
\end{equation}
The quantity $a^2-b$ must be a perfect square if the de-nesting is to succeed.  If so, call its square root $r $.
Now using the uniqueness of quadratic sure, square (1) and equate rational parts, getting:
\begin{equation} \tag{4}
a
=x + y.
\end{equation}
Equations (3) and (4) constitute a linear system for $x$ and $y$, which is easily solved, thus
\begin{equation*}
x = \frac{a + r}{2}
\qquad \text{and} \qquad
y = \frac{a - r}{2}.
\end{equation*}
Plug these results into the given problem  ($a=6, b=20$) and infer that the de-nested roots are:
$$\sqrt {6+2\sqrt {5}} =\sqrt {5}+1
\qquad \text{and} \qquad
\sqrt {6-2\sqrt {5}} =\sqrt {5}-1
$$
with the difference of $2$.
Note:If in this development we allow negative values of $b $ and drop the rationality requirement on $r$, this is also a method to extract square roots of complex numbers.
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