now finding is straightforward. to find substitute: then and thus
finally let and to get: note that now everything is nice and easy and you're good to go!
Remark: i think this is true that for any we have where is some rational function of .
actually i think where both and are monic polynomials with integer coefficients! for example, in your problem: i also think that we
always have these claims are either wrong or easy to prove! (Rofl) ... let's just leave it as a challenge for whoever has time and patience to try it.
by the way, the identity shows that my claims are not based on nothing! (Evilgrin)
My solution is to substitute
Although it looks very troublesome , it become quite well after separating it into fractions .
However , i think your method is better than what i am using . (Clapping)