Please answer qno 7i . Quick.

Dear Student, 

  We have, 
               $\frac{{\left(27\right)}^{{\displaystyle \frac{2n}{3}}}\times {\left(8\right)}^{{\displaystyle \frac{-n}{6}}}}{{\left(18\right)}^{{\displaystyle \frac{-n}{2}}}}$
Solving, we get, 

                    $$\begin{gathered} \frac{{\left(27\right)}^{{\displaystyle \frac{2n}{3}}}\times {\left(8\right)}^{{\displaystyle \frac{-n}{6}}}}{{\left(18\right)}^{{\displaystyle \frac{-n}{2}}}} \\ \to \frac{{\left[\right(3)}^3{]}^{{\displaystyle \frac{2n}{3}}}\times {\left[\right(2)}^3{]}^{{\displaystyle \frac{-n}{6}}}}{{\left(18\right)}^{{\displaystyle \frac{-n}{2}}}} \\ \to \frac{{\left(3\right)}^{2n}\times {\left(2\right)}^{{\displaystyle \frac{-n}{2}}}}{{\left(18\right)}^{{\displaystyle \frac{-n}{2}}}} (\because {\left[{\left(a\right)}^m\right]}^n={\left(a\right)}^{m\times n} ) \\ \to {{\left(3\right)}^{2n}\times \left(\frac{\left(2\right)}{\left(18\right)}\right)}^{\frac{-n}{2}} (\because \frac{a^m}{b^m}={\left(\frac{a}{b}\right)}^m ) \\ \to {\left(3\right)}^{2n}\times {\left(\frac{1}{9}\right)}^{\frac{-n}{2}} \\ \to {\left(3\right)}^{2n}\times {\left(9\right)}^{\frac{n}{2}} (\because {\left(\frac{1}{a}\right)}^{-m}={\left(a\right)}^m ) \\ \to {\left(3\right)}^{2n}\times {\left[{\left(3\right)}^2\right]}^{\frac{n}{2}} \\ \to {\left(3\right)}^{2n}\times {\left(3\right)}^n \\ \to {\left(3\right)}^{2n+n} (\because {\left(a\right)}^m\times {\left(a\right)}^n={\left(a\right)}^{m+n }) \\ \to {\left(3\right)}^{3n} or {\left(27\right)}^n \end{gathered}$$

Hope this information clears your doubts.

Regards