<s>[INST] Simplifying fractions How to transform $$\left({1+\sqrt{1-4ab}\over2a}\right)^n-\left({1-\sqrt{1-4ab}\over2a}\right)^n\over \left({1+\sqrt{1-4ab}\over2a}\right)^m-\left({1-\sqrt{1-4ab}\over2a}\right)^m$$ into $$\left({b\over a}\right)^n -1\over \left({b\over a}\right)^m -1$$?
Thanks.
Added: As per comments, there is an additional assumption that $a,b\gt 0$, $b=1-a$, and $b\gt a$.
 [/INST] With the added information, which is key (otherwise, the two expressions are just not equal in general, as noted by Ross):
If $b=1-a$ and $b\gt a$, then $1-a\gt a$ so $1\gt 2a$. Then
$$1 -4ab = 1-4a(1-a) = 1-4a+4a^2 = (1-2a)^2,$$
and since $1-2a\gt 0$, we have
$$\sqrt{1-4ab} = \sqrt{(1-2a)^2} = |1-2a| = 1-2a.$$
So
$$\begin{align*}
\frac{1+\sqrt{1-4ab}}{2a} &= \frac{1+(1-2a)}{2a} = \frac{2(1-a)}{2a} = \frac{b}{a},\\
\frac{1-\sqrt{1-4ab}}{2a} &= \frac{1-(1-2a)}{2a} = \frac{2a}{2a} = 1.
\end{align*}$$
That means that
$$\begin{align*}
\left({1+\sqrt{1-4ab}\over2a}\right)^n-\left({1-\sqrt{1-4ab}\over2a}\right)^n\over \left({1+\sqrt{1-4ab}\over2a}\right)^m-\left({1-\sqrt{1-4ab}\over2a}\right)^m & =
\frac{\left(\frac{b}{a}\right)^n - 1^n}{\left(\frac{b}{a}\right)^m - 1^m}\\
&=\frac{\left(\frac{b}{a}\right)^n - 1}{\left(\frac{b}{a}\right)^m - 1}.
\end{align*}$$
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