If the local maximum value of the function $\mathrm{f}\left(\mathrm{x}\right)={\left(\frac{\sqrt{3\mathrm{e}}}{2\mathrm{sinx}}\right)}^{{\sin }^2\mathrm{x}},\mathrm{x}\in \left(0,\frac{\mathrm{\pi }}{2}\right)$ is $\frac{\mathrm{k}}{\mathrm{e}},$ then ${\left(\frac{\mathrm{k}}{\mathrm{e}}\right)}^8+\frac{{\mathrm{k}}^8}{{\mathrm{e}}^5}+{\mathrm{k}}^8$ is equal to 

Given, $\mathrm{f}\left(\mathrm{x}\right)={\left(\frac{\sqrt{3\mathrm{e}}}{2\mathrm{sinx}}\right)}^{{\sin }^2\mathrm{x}},\mathrm{x}\in (0,\frac{\mathrm{\pi }}{2})$
Let $\left(\frac{{\displaystyle \sqrt{3\mathrm{e}}}}{{\displaystyle 2\mathrm{sinx}}}\right)=\mathrm{t},$
Consider $\mathrm{y}={\mathrm{t}}^{\frac{3\mathrm{e}}{4{\mathrm{t}}^2}}$
for Max $\frac{\mathrm{dy}}{\mathrm{dt}}={\mathrm{t}}^{\frac{3\mathrm{e}}{4{\mathrm{t}}^2}}[\frac{1}{\mathrm{t}}\left(\frac{3\mathrm{e}}{4{\mathrm{t}}^2}\right)+\left(\frac{-6\mathrm{e}}{4{\mathrm{t}}^3}\right)\mathrm{lnt}]=0$
$$\begin{gathered} \Rightarrow \frac{3\mathrm{e}}{4{\mathrm{t}}^3}-\frac{6\mathrm{e}}{4{\mathrm{t}}^3}\ln \mathrm{t}=0 \\ \Rightarrow \mathrm{lnt}=\frac{1}{2}\Rightarrow \mathrm{t}={\mathrm{e}}^{1/2} \\ \therefore \mathrm{y}={\left({\mathrm{e}}^{1/2}\right)}^{\frac{3\mathrm{e}}{4\mathrm{e}}}={\mathrm{e}}^{3/8}=\frac{\mathrm{k}}{\mathrm{e}}\Rightarrow {\mathrm{k}}^8={\mathrm{e}}^{11} \\ \mathrm{Now}, {\left(\frac{\mathrm{k}}{\mathrm{e}}\right)}^8+\frac{{\mathrm{k}}^8}{{\mathrm{e}}^5}+{\mathrm{k}}^8 \\ =\frac{{\mathrm{e}}^{11}}{{\mathrm{e}}^8}+\frac{{\mathrm{e}}^{11}}{{\mathrm{e}}^5}+{\mathrm{e}}^{11}={\mathrm{e}}^3+{\mathrm{e}}^6+{\mathrm{e}}^{11} \end{gathered}$$