Question

Show that:
$\begin{array}{cc}& {\left(\frac{a^m}{a^{-n}}\right)}^{m-n}\times {\left(\frac{a^n}{a^{-l}}\right)}^{n-l}\times {\left(\frac{a^l}{a^{-m}}\right)}^{l-m}\\ & =1\end{array}$

Answer 1

Step : Simplifying the LHS of given expression
Given expression:
${\left(\frac{a^m}{a^{-n}}\right)}^{m-n}\times {\left(\frac{a^n}{a^{-l}}\right)}^{n-l}\times {\left(\frac{a^l}{a^{-m}}\right)}^{l-m}=1$
Taking L.H.S of given expression, we have:
$\begin{array}{cc}& {\left(\frac{a^m}{a^{-n}}\right)}^{m-n}\times {\left(\frac{a^n}{a^{-l}}\right)}^{n-l}\times {\left(\frac{a^l}{a^{-m}}\right)}^{l-m}\\ & ={(a^m\times a^n)}^{m-n}\times {(a^n\times a^l)}^{n-l}\times {(a^l\times a^m)}^{l-m}\\ & (\because \frac{1}{a^{-x}}=a^x)\\ & ={\left(a^{m+n}\right)}^{(m-n)}\times {\left(a^{n+l}\right)}^{(n-l)}\times {\left(a^{l+m}\right)}^{(l-m)}\\ & [\because a^m\times a^n=a^{m+n}]\end{array}$
$\begin{array}{cc}& =a^{(m+n)(m-n)}\times a^{(n+l)(n-l)}\times a^{(l+m)(l-m)}\\ & =a^{m^2-n^2}\times a^{n^2-l^2}\times a^{l^2-m^2}\\ & =a^{m^2-n^2+n^2-l^2+l^2-m^2}\\ & [\because a^x\times a^y\times a^z=a^{x+y+z}]\\ & =a^0=1\end{array}$
Hence proved.