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

$\text { Let } y=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^2 x}$ <br/><br/>$$ \ln \mathrm{y}=\sin ^2 \mathrm{x} \cdot \ln \left(\frac{\sqrt{3 \mathrm{e}}}{2 \sin \mathrm{x}}\right) $$ <br/><br/>$$ \frac{1}{y} y^{\prime}=\ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) 2 \sin x \cos x+\sin ^2 x \frac{2 \sin x}{\sqrt{3 e}} \frac{\sqrt{3 e}}{2}(-\operatorname{cosec} x \cot x) $$ <br/><br/>For maxima or minima, $\frac{d y}{d x}=0$ <br/><br/>$$ \Rightarrow \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) 2 \sin x \cos x-\sin x \cos x=0 $$ <br/><br/>$$ \Rightarrow \sin x \cos x\left[2 \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right)-1\right]=0 $$ <br/><br/>$\Rightarrow \ln \left(\frac{3 \mathrm{e}}{4 \sin ^2 \mathrm{x}}\right)=1$ <br/><br/>$\Rightarrow \frac{3 e}{4 \sin ^2 x}=e$ <br/><br/>$\Rightarrow \sin ^2 x=\frac{3}{4}$ <br/><br/>$$ \Rightarrow \sin \mathrm{x}=\frac{\sqrt{3}}{2} \quad\left(\text { as } \mathrm{x} \in\left(0, \frac{\pi}{2}\right)\right) $$ <br/><br/>$$ \Rightarrow \text { Local max value }=\left(\frac{\sqrt{3 \mathrm{e}}}{\sqrt{3}}\right)^{3 / 4}=\mathrm{e}^{3 / 8}=\frac{\mathrm{k}}{\mathrm{e}} $$ <br/><br/>$\Rightarrow \mathrm{k}^8=\mathrm{e}^{11}$ <br/><br/>$\therefore$ $\left(\frac{k}{e}\right)^8+\frac{k^8}{e^5}+k^8=e^3+e^6+e^{11}$