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

Given function is $f\left(x\right)={\left(\frac{\sqrt{3}e}{2\sin x}\right)}^{{\sin }^{2}x}$

For maxima or minima $f^{\text{'}}\left(x\right)=0$

$f^{\text{'}}\left(x\right)=f\left(x\right)\left[2\sin x\cos x\times \ln \left(\frac{\sqrt{3}\mathrm{e}}{2\sin x}\right)+{\sin }^{2}x\frac{1\times 2\sin x}{\sqrt{3}e}\times \frac{\sqrt{3}e}{2}\left(-\frac{1}{{\sin }^{2}x}\times \cos x\right)\right]$

$\Rightarrow f\left(x\right)\left[\sin 2x\ln \left(\frac{\sqrt{3}\mathrm{e}}{2\sin x}\right)-\sin x\cos x\right]=0$

Now on equation we get, $\sin 2x=0$ (not possible)

So, $\ln \left(\frac{\sqrt{3}\mathrm{e}}{2\sin x}\right)=+\frac{1}{2}$

$\Rightarrow \frac{\sqrt{3}\times \sqrt{e}}{2\sin x}=e^{\frac{1}{2}}$

$\Rightarrow \sin x=\frac{\sqrt{3}}{2}$

$\Rightarrow f_{\max }={\left(e\right)}^{\frac{3}{8}}=\frac{e^{{\displaystyle \frac{11}{8}}}}{e}\Rightarrow k=e^{\frac{11}{8}}$

$\Rightarrow {\left(\frac{k}{e}\right)}^8+\frac{k^8}{e^5}+k^8=e^3+e^6+e^{11}$