Question:
Show that
$\frac{\left(a+\frac{1}{b}\right)^{m} \times\left(a-\frac{1}{b}\right)^{n}}{\left(b+\frac{1}{a}\right)^{m} \times\left(b-\frac{1}{a}\right)^{n}}=\left(\frac{a}{b}\right)^{m+n}$
Solution:
$=\frac{\left(\frac{a b+1}{b}\right)^{m} \times\left(\frac{a b-1}{b}\right)^{n}}{\left(\frac{a b+1}{a}\right)^{m} \times\left(\frac{a b+1}{a}\right)^{n}}$
$=\left(\frac{a}{b}\right)^{m} \times\left(\frac{a}{b}\right)^{n}$
$=\left(\frac{a}{b}\right)^{m+n}$
Hence, LHS = RHS