Question

a
$( \frac{ {a}^{m} }{ {a}^{ - n} } ) {}^{m - n} \times ( \frac{ {a}^{n} }{ {a}^{ - p} }) ^{n - p} \times ( \frac{ {a}^{p} }{ {a}^{ - m} }) {}^{p - m} = 1$
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Answer 1

To prove :

$\sf \: \bigg( \dfrac{ {a}^{m} }{ {a}^{ - n} } \bigg)^{(m - n)} \: \times \: \bigg( \dfrac{ {a}^{n} }{ {a}^{ - p} } \bigg)^{(n - p)} \: \times \: \bigg( \dfrac{ {a}^{p} }{ {a}^{ -m} } \bigg)^{(p - m )} \: = 1$

Identity used :

$\bullet \: \: \sf\dfrac{a^x}{a^y} \: = \: a^{(x - y)}\\\\\bullet\:\:\sf{(a^x)^y \: = \: a^{xy}}\\\\\bullet\:\:\sf(a^x) \times (a^y) \: = \: a^{x + y} \\ \\ \sf \bullet \: \: \: {a}^{0} = \: 1 \\ \\ \sf \bullet \: \: ( {x}^{2} - {y}^{2} ) = (x + y)(x - y)$

Solution :

$\sf \: \bigg( \dfrac{ {a}^{m} }{ {a}^{ - n} } \bigg)^{(m - n)} \: \times \: \bigg( \dfrac{ {a}^{n} }{ {a}^{ - p} } \bigg)^{(n - p)} \: \times \: \bigg( \dfrac{ {a}^{p} }{ {a}^{ -m} } \bigg)^{(p - m )} \: = 1 \\ \\ \sf Now, \:\:\\ \\ \sf \implies \: LHS \: \: = \: \bigg( {a}^{m - ( - n)} \: \bigg)^{(m - n)} \: \times \: \bigg( \: {a}^{n - ( - p)} \: \bigg)^{(n - p)} \: \times \: \bigg( \: {a}^{p - ( - m)} \: \bigg)^{(p - m )} \: \\ \\ \\ \sf \implies \: LHS \: \: = \: \bigg( {a}^{(m + n)} \: \bigg)^{(m - n)} \: \times \: \bigg( \: {a}^{(n + p)} \: \bigg)^{(n - p)} \: \times \: \bigg( \: {a}^{(p + m)} \: \bigg)^{(p - m )} \: \\ \\ \sf \implies \: LHS \: \: = \: {a}^{(m + n)(m - n)} \: \times \: {a}^{( n + p)(n - p)} \: \times \: {a}^{(p + m)(p - m)} \\ \\ \sf \implies \: \: LHS \: \: = \: \: {a}^{( {m}^{2} - {n}^{2} ) } \: \times \: {a}^{( {n}^{2} - {p}^{2} ) } \: \times \: {a}^{( {p}^{2} - {m}^{2} ) } \\ \\ \sf \implies \: \: LHS \: \: = \: \: {a}^{ \: [ \: ( {m}^{2} - {n}^{2} ) \: + \: ( {n}^{2} - {p}^{2} ) \: + \:( {p}^{2} - {m}^{2} ) \: \: ] } \: \\ \\ \sf \implies \: \: LHS \: \: = \: {a}^{ \: [ \: {m}^{2} - {n}^{2} \: + \: {n}^{2} - {p}^{2} \: + \:{p}^{2} - {m}^{2} \: \: ] } \\ \\ \sf \implies \: \: LHS \: \: = \: \: {a}^{0} \\ \\ \sf \implies \: \: LHS \: \: = \:1$

Also ,

RHS = 1

This gives , LHS = RHS

Hence proved!!