How to prove an identity in radicals? (4 / (3 - sqrt(5))) ^ 2 - ((6 - 5 * sqrt(6)) / (5 - sqrt(6))) ^ 2 = 2 * sqrt(61 + 24*sqrt(5))
$$\left(\frac{4}{3-\sqrt{5}}\right)^2 - \left(\frac{6 - 5 \sqrt{6}}{5 - \sqrt{6}}\right)^2 = 2\sqrt{61+24\sqrt{5}}$$
How to prove it is right equality?
I come up with $\dfrac{16}{14-6\sqrt{5}}-6=2\sqrt{61+24\sqrt{5}}$, but still can't get it to the obvious equality.
Any ideas?

Trivial. Lets simplify the left part:
multiply first fraction to the sum of expressions in its denominator, and get whole part in second fraction and do with second fraction the same that we did with first:
$(\frac{4(3+\sqrt{5})}{9-5})^2 - (\frac{25-19 - \sqrt{6}}{5-\sqrt{6}}) = (3+\sqrt{5})^2 - (5-\frac{19(5+\sqrt{6})}{25-6})^2 = (3+\sqrt{5})^2 - 6 = 8 + 6\sqrt{5}$
Now lets prove this:
$8+6\sqrt{5}=2\sqrt{61+24\sqrt{5}}$
divide by two and square:
$4+3\sqrt{5} = \sqrt{61+24\sqrt{5}}$
$16+45+24\sqrt{5}=61+24\sqrt{5}$
Thats all.
