Simplify this expression with nested radical signs My question is-
Simplify:
$$\frac1{\sqrt{12-2\sqrt{35}}}-\frac2{\sqrt{10+2\sqrt{21}}}-\frac1{\sqrt{8+2\sqrt{15}}}$$
 A: The general method for going from a square root expression to a polynomial equation is as follows:
Rationalize the denominators and then set your expression equal to $x$.
Now, isolate one of the square roots on one side of the equation and square everything. Repeating this will eventually remove all square roots and will give you a polynomial for $x$.
Then, use estimates for $x$ to decide which root of the polynomial is $x$.

In that particular case, the answer is $x=\sqrt 3$, and we can do much better by recognizing the general structure.
The rationalized expression is actually of the form:
$$\frac{\sqrt{b+c+2\sqrt{bc}}}2-\frac{\sqrt{a+c-2\sqrt{ac}}}2-\frac{\sqrt{a+b-2\sqrt{ab}}}2$$
for $a=3$, $b=5$ and $c=7$.
These are complete squares, so we get:
$$\frac 12 (\sqrt b +\sqrt c +\sqrt a-\sqrt c +\sqrt a -\sqrt b) = \sqrt a=\sqrt 3.$$
A: Let's follow this elementary way to work out things:
$$\frac1{\sqrt{12-2\sqrt{35}}}-\frac2{\sqrt{10+2\sqrt{21}}}-\frac1{\sqrt{8+2\sqrt{15}}}= \frac1{\sqrt{({\sqrt{7}-\sqrt{5}})^2}}-\frac2{\sqrt{({\sqrt{7}+\sqrt{3}})^2}}-\frac1{\sqrt{({\sqrt{5}+\sqrt{3}})^2}}=\frac1{({\sqrt{7}-\sqrt{5}})}-\frac2{({\sqrt{7}+\sqrt{3}})}-\frac1{({\sqrt{5}+\sqrt{3}})}=$$
$$\frac{\sqrt{7}+\sqrt{5}-\sqrt{7}+\sqrt{3}-\sqrt{5}+\sqrt{3}}{2} =\sqrt3.$$
The proof is complete. 
A: Hint: Rationalize the denominators:
For example:
$$\sqrt{12-2\sqrt{35}}\sqrt{12+2\sqrt{35}}=\sqrt{144-4\cdot35}=\sqrt{4}=2$$
A: $$
\begin{align} & {}\quad \frac1{\sqrt{12-2\sqrt{35}}}-\frac2{\sqrt{10+2\sqrt{21}}}-\frac1{\sqrt{8+2\sqrt{15}}}\\[10pt]
& =\frac {1}{\sqrt{ 12-2 \sqrt {35}}} \frac {\sqrt{ 12+2 \sqrt {35}}}{\sqrt{ 12+2 \sqrt {35}}}- \frac {2 }{\sqrt{ 10+2 \sqrt {21}}} \frac {\sqrt{ 10-2 \sqrt {21}}}{\sqrt{ 10-2 \sqrt {21}}}- \frac{1}{\sqrt {8+2\sqrt {15}}}\frac{\sqrt {8-2\sqrt {15}}}{\sqrt {8-2\sqrt {15}}}\\[10pt]
& =\frac {\sqrt{ 12+2 \sqrt {35}}}{2}-\frac {\sqrt{ 10-2 \sqrt {21}}}{2}-\frac {\sqrt {8-2\sqrt {15}}}{2}\\[10pt]
& =\frac {{\sqrt{ 12+2 \sqrt {35}}}- {\sqrt{ 10-2 \sqrt {21}}}- {\sqrt {8-2\sqrt {15}}}}{2}\\[10pt]
& = \frac {{{|\sqrt 5 + \sqrt 7|}}- {{|\sqrt 3 - \sqrt 7|}}- {{|\sqrt 3 - \sqrt 5|}}}{2}=\sqrt 3 
\end{align}
$$
A: If you did not notice that each radical denests via $\rm\:a+b+2\sqrt{ab}\, =\, (\sqrt{a}+\sqrt{b})^2\:$ then you could easily calculate this by using this radical denesting formula that I discovered as a teenager.

Simple Denesting Rule $\rm\ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \  \color{brown}{divide\ out}\ \sqrt{trace} $
Recall $\rm\: w = a + b\sqrt{n}\: $ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2 $  
and, furthermore,  $\rm\:w\:$ has trace $\rm\: =\: w+w' =  (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\:  2\:a$

Here $\rm\:a\!+\!b+\sqrt{ab}\:$ has norm $\rm\:(a\!-\!b)^2.\:$ $\rm\ \color{blue}{subtracting\ out}\ \sqrt{norm}\ = a\!-\!b\ $ yields $\rm\:  2b+\sqrt{ab}\:$
and this has $\rm\ \sqrt{trace}\: =\: 2\sqrt{b},\ \ hence,\ \  \ \color{brown}{dividing\ it\ out}\ $ of this yields the sqrt: $\rm\ \sqrt{b}+\sqrt{a}.$
See this answer for general radical denesting algorithms.
