Question

Please help me by solving the problem ​

Answer 1

Solution -

$\sf{ ( \dfrac{ a^m }{ a ^ {-n} } ) ^ { m - n } \times ( \dfrac{ a^n }{ a ^ {-l} } ) ^ { n - l } \times ( \dfrac{ a^l }{ a ^ {-m} } ) ^ { l - m } } \\ \\ \sf{ \implies { ( a ^ { m + n } ) ^ { m - n } \times ( a ^ { n + l} ) ^ { n - l } \times ( a ^ { l + m } ) ^ { l - m } }} \\ \\ \sf{ \implies { a ^ { ( m^2 - n ^ 2 ) } \times a ^ { ( n ^ 2 - l ^ 2 ) } \times a ^ { ( l ^ 2 - m ^ 2) } }} \\ \\ \sf{ \implies { a ^ { ( m ^2 - n^2 + n^2 - l^2 + l^2 - m ^ 2 ) } }} \\ \\ \sf{ \implies { a ^ 0 }} \\ \\ \sf{ \implies { 1 }} \\ \\ \sf{ \bold { Hence \: Shown }}$

AddiTiOnaL InFoRmAtIon -

$\sf{ a ^ 0 = 1 }$

$\sf{ a ^ m \times a ^ n = a ^ { m + n } }$

$\sf{ \dfrac{ a ^ m }{ a ^ n } = a ^ { m - n}}$

__________________

Answer 2

$\huge\sf\blue{Given}$

✭ $\sf\bigg\lgroup\dfrac{a^m}{a^{-n}} \bigg\rgroup^{m-n} \times {\bigg\lgroup \dfrac{a^n}{a^{-l}} \bigg\rgroup}^{n-1} \times \bigg\lgroup \dfrac{a^l}{a^{-m}} \bigg\rgroup = 1$

$\rule{110}1$

$\huge\sf\gray{To\;Prove}$

◈ The above statement?

$\rule{110}1$

$\huge\sf\purple{Steps}$

$\begin{lgathered}\sf\bigg\lgroup\dfrac{a^m}{a^{-n}} \bigg\rgroup^{m-n} \times {\bigg\lgroup \dfrac{a^n}{a^{-l}} \bigg\rgroup}^{n-1} \times \bigg\lgroup \dfrac{a^l}{a^{-m}} \bigg\rgroup = 1 \\ \\ \sf{ \dashrightarrow { ( a ^ { m + n } ) ^ { m - n } \times ( a ^ { n + l} ) ^ { n - l } \times ( a ^ { l + m } ) ^ { l - m } }} \\ \\ \sf{ \dashrightarrow{ a ^ { ( m^2 - n ^ 2 ) } \times a ^ { ( n ^ 2 - l ^ 2 ) } \times a ^ { ( l ^ 2 - m ^ 2) } }} \\ \\ \sf{ \dashrightarrow { a ^ { ( m ^2 - n^2 + n^2 - l^2 + l^2 - m ^ 2 ) } }} \\ \\ \sf{ \dashrightarrow{ a ^ 0 }} \\ \\ \sf{ \dashrightarrow { 1 }} \\ \\ \sf{ \sf { Hence \ Proved!!}}\end{lgathered}$

$\rule{170}3$