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?

Simplify both sides to $8+6\sqrt{5}$, as they are both equal to this.
Just square the RHS to see this and rationalise the denominators on the LHS.
After the rationalisation of the denominators on the LHS (which is very quick) you obtain $(3+\sqrt{5})^2-6,$ and hence the $8+6\sqrt{5}.$
