Question

Prove that ${\left(\frac{x^l}{x^{-m}}\right)}^{l-m}\times {\left(\frac{x^m}{x^{-n}}\right)}^{m-n}\times {\left(\frac{x^n}{x^{-l}}\right)}^{n-l}=1$

Answer 1

To prove: ${\left(\frac{x^l}{x^{-m}}\right)}^{l-m}\times {\left(\frac{x^m}{x^{-n}}\right)}^{m-n}\times {\left(\frac{x^n}{x^{-l}}\right)}^{n-l}=1$

$\begin{array}{l}={\left(\frac{x^l}{x^{-m}}\right)}^{l-m}\times {\left(\frac{x^m}{x^{-n}}\right)}^{m-n}\times {\left(\frac{x^n}{x^{-l}}\right)}^{n-l}\\ ={\left(x^{l+m}\right)}^{l-m}\times {\left(x^{m+n}\right)}^{m-n}\times {\left(x^{n+l}\right)}^{n-l}\\ =\left(x^{(l+m)\left(l-m\right)}\right)\times \left(x^{(m+n)\left(m-n\right)}\right)\times \left(x^{(n+l)\left(n-l\right)}\right)\left(\because {\left(a^m\right)}^n=a^{mn}\right)\\ =\left(x^{l^2-m^2}\right)\times \left(x^{m^2-n^2}\right)\times \left(x^{n^2-l^2}\right)\left(\because \left(a+b\right)\left(a-b\right)=a^2-b^2\right)\\ =x^{l^2-m^2+m^2-n^2+n^2-l^2}(\because a^m\times a^n=a^{m+n})\\ =x^0\\ =1\end{array}$

Hence, it is proved that ${\left(\frac{x^l}{x^{-m}}\right)}^{l-m}\times {\left(\frac{x^m}{x^{-n}}\right)}^{m-n}\times {\left(\frac{x^n}{x^{-l}}\right)}^{n-l}=1$