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

We need to find the maximum value of $$f(x) = \left(\frac{\sqrt{3e}}{2\sin x}\right)^{\sin^2 x}$$ for $$x \in \left(0, \frac{\pi}{2}\right)$$.

Let $$g(x) = \ln f(x) = \sin^2 x \cdot \ln\left(\frac{\sqrt{3e}}{2\sin x}\right)$$.

Let $$t = \sin x$$ where $$t \in (0, 1)$$. Then:

$$ g = t^2 \left[\frac{1}{2}\ln(3e) - \ln(2t)\right] = t^2 \left[\frac{1}{2}\ln 3 + \frac{1}{2} - \ln 2 - \ln t\right] $$

Differentiating with respect to $$t$$:

$$ \frac{dg}{dt} = 2t\left[\frac{1}{2}\ln(3e) - \ln(2t)\right] + t^2 \cdot \left(-\frac{1}{t}\right) $$

$$ = t\left[\ln\left(\frac{3e}{4t^2}\right) - 1\right] $$

Setting $$\frac{dg}{dt} = 0$$ (with $$t > 0$$):

$$ \ln\left(\frac{3e}{4t^2}\right) = 1 \implies \frac{3e}{4t^2} = e \implies t^2 = \frac{3}{4} \implies t = \frac{\sqrt{3}}{2} $$

This corresponds to $$\sin x = \frac{\sqrt{3}}{2}$$, i.e., $$x = \frac{\pi}{3}$$.

At $$t = \frac{\sqrt{3}}{2}$$:

$$ f = \left(\frac{\sqrt{3e}}{2 \cdot \frac{\sqrt{3}}{2}}\right)^{3/4} = \left(\frac{\sqrt{3e}}{\sqrt{3}}\right)^{3/4} = \left(\sqrt{e}\right)^{3/4} = e^{3/8} $$

So the maximum value of $$f(x)$$ is $$\frac{k}{e} = e^{3/8}$$, giving $$k = e^{11/8}$$.

$$\left(\frac{k}{e}\right)^8 = \left(e^{3/8}\right)^8 = e^3$$

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

$$k^8 = \left(e^{11/8}\right)^8 = e^{11}$$

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

The correct answer is Option A: $$e^3 + e^6 + e^{11}$$.