Question

\[ \frac{(a+\frac{1}{b})^m (a-\frac{1}{b})^n}{(b+\frac{1}{a})^m (b-\frac{1}{a})^n} \]

Solution

\[ a+\frac{1}{b} = \frac{ab+1}{b},\quad a-\frac{1}{b} = \frac{ab-1}{b} \] \[ b+\frac{1}{a} = \frac{ab+1}{a},\quad b-\frac{1}{a} = \frac{ab-1}{a} \] \[ = \frac{\left(\frac{ab+1}{b}\right)^m \left(\frac{ab-1}{b}\right)^n}{\left(\frac{ab+1}{a}\right)^m \left(\frac{ab-1}{a}\right)^n} \] \[ = \frac{(ab+1)^m (ab-1)^n}{b^{m+n}} \cdot \frac{a^{m+n}}{(ab+1)^m (ab-1)^n} \] \[ = \frac{a^{m+n}}{b^{m+n}} \] \[ = \left(\frac{a}{b}\right)^{m+n} \]

Answer

\[ \boxed{\left(\frac{a}{b}\right)^{m+n}} \]

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