Question

Show that:

$\frac{{(a+\frac{1}{b})}^m\times {(a-\frac{1}{b})}^n}{{(b+\frac{1}{a})}^m\times {(b-\frac{1}{a})}^n}={\left(\frac{a}{b}\right)}^{m+n}$

Answer 1

Given: $\frac{{(a+\frac{1}{b})}^m\times {(a-\frac{1}{b})}^n}{{(b+\frac{1}{a})}^m\times {(b-\frac{1}{a})}^n}={\left(\frac{a}{b}\right)}^{m+n}$

$LHS$

$=\frac{{(a+\frac{1}{b})}^m\times {(a-\frac{1}{b})}^n}{{(b+\frac{1}{a})}^m\times {(b-\frac{1}{a})}^n}$

$=\frac{{\left(\frac{ab+1}{b}\right)}^m\times {\left(\frac{ab-1}{b}\right)}^n}{{\left(\frac{ab+1}{a}\right)}^m\times {\left(\frac{ab-1}{a}\right)}^n}$

$=\frac{\frac{(ab+1{)}^m}{b^m}\times \frac{(ab-1{)}^n}{b^n}}{\frac{(ab+1{)}^m}{a^m}\times \frac{(ab-1{)}^n}{a^n}}$

$=\frac{(ab+1{)}^m}{(ab+1{)}^m}\times \frac{(ab-1{)}^n}{(ab-1{)}^n}\times \frac{a^m\times a^n}{b^m\times b^n}$

$=\frac{a^{m+n}}{b^{m+n}}$

$={\left(\frac{a}{b}\right)}^{m+n}$

$=RHS$

Hence, proved.