Question

please solve this for me
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Answer 1

hence the prove.

hope you understand.

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Answer 2

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

L.H.S = R.H.S

Hence , Proved.