<s>[INST] 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?
 [/INST] You want to show:
$$\left(\frac{4}{3-\sqrt{5}}\right)^2 - \left(\frac{6 - 5 \sqrt{6}}{5 - \sqrt{6}}\right)^2 = 2\sqrt{61+24\sqrt{5}}$$
First multiply out the squares
$$\frac{16}{14-6\sqrt{5}} - \frac{186-60\sqrt{6}}{31-10\sqrt{6}} = 2\sqrt{61+24\sqrt{5}}$$
Then simplify it
$$\frac{16}{14-6\sqrt{5}} - 6 = 2\sqrt{61+24\sqrt{5}}$$
now we can start getting rid of the square roots on the right hand side by squaring both sides
$$\frac{256}{376-168\sqrt{5}} - \frac{192}{14-6\sqrt{5}} + 36 = 244 + 96 \sqrt{5}$$
at this point it makes sense to rationalize the denominators of the fractions
$$\left(\frac{256}{376-168\sqrt{5}}\right)\left(\frac{376+168\sqrt{5}}{376+168\sqrt{5}}\right) - \left(\frac{192}{14-6\sqrt{5}}\right)\left(\frac{14+6\sqrt{5}}{14+6\sqrt{5}}\right) + 36 = 244 + 96 \sqrt{5}$$
which simplifies to
$$376 + 168\sqrt{5} - 168 - 72\sqrt{5} + 36 = 244 + 96 \sqrt{5}$$
collecting like terms now gives
$$(376 - 168 + 36) + (168 - 72)\sqrt{5} = 244 + 96 \sqrt{5}$$
which is easily seem to be equal.

Another approach is, using the fact that algebraic numbers cannot be too close together - compute the first few digits of both sides of the original identity and compare them.
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