If the total maximum value of the function f(x) = ( 3e 2 x ) ^ ^ 2...

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

For maxima or minima ${f}^{'} ( x ) = 0$

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

$⇒ f\left(x\right)\left[\sin 2xln\left(\dfrac{\sqrt{3}e}{2\sin x}\right) - \sin x \cos x\right] = 0$

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

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

$⇒ \dfrac{\sqrt{3} \times \sqrt{e}}{2\sin x} = e\dfrac{1}{2}$

$⇒ \sin x = \dfrac{\sqrt{3}}{2}$

⇒ {f}_\max = \left(e\right)\dfrac{3}{8} = \dfrac{{e}^{\dfrac{11}{8}}}{e} ⇒ k = e\dfrac{11}{8}

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