Let $$[\cdot]$$ denote the greatest integer function. If $$\int_0^{e^3} \left[\frac{1}{e^{x-1}}\right] dx = \alpha - \log_e 2$$, then $$\alpha^3$$ is equal to _____.

The integrand can be rewritten in exponential form:

$$\frac{1}{e^{x-1}} = e^{-(x-1)} = e^{1-x}.$$

Therefore the integral becomes

$$\int_{0}^{e^{3}} \bigl[e^{1-x}\bigr] \, dx.$$

Because $$e^{1-x}$$ is a strictly decreasing function of $$x$$, let us first see for which part of the interval $$[0,e^{3}]$$ the value of $$e^{1-x}$$ is $$\ge 1$$.

Set $$e^{1-x} \ge 1 \iff 1 - x \ge 0 \iff x \le 1.$$
Thus for all $$x > 1$$, $$e^{1-x} \lt 1$$ and hence $$\bigl[e^{1-x}\bigr] = 0.$$

So the integral contributes only over $$[0,1]$$:

$$\int_{0}^{e^{3}} \bigl[e^{1-x}\bigr] \, dx = \int_{0}^{1} \bigl[e^{1-x}\bigr] \, dx + \int_{1}^{e^{3}} 0 \, dx = \int_{0}^{1} \bigl[e^{1-x}\bigr] \, dx.$$

Next find how $$e^{1-x}$$ ranges on $$[0,1]$$.
At $$x = 0$$: $$e^{1-0} = e.$$
At $$x = 1$$: $$e^{1-1} = 1.$$

Thus $$e^{1-x}$$ decreases from $$e \approx 2.718$$ down to $$1$$. Its greatest-integer (floor) values can therefore only be $$2$$ or $$1$$.

Find the point where $$e^{1-x} = 2$$:

$$e^{1-x} = 2 \;\Longrightarrow\; 1 - x = \ln 2 \;\Longrightarrow\; x = 1 - \ln 2.$$

So we split the integral:

$$\int_{0}^{1} \bigl[e^{1-x}\bigr]\,dx = \int_{0}^{\,1-\ln 2} 2 \, dx \;+\; \int_{\,1-\ln 2}^{1} 1 \, dx.$$

Compute each part:

First part: $$\int_{0}^{\,1-\ln 2} 2\,dx = 2\bigl(1 - \ln 2 - 0\bigr) = 2 - 2\ln 2.$$

Second part: $$\int_{\,1-\ln 2}^{1} 1\,dx = 1\bigl(1 - (1-\ln 2)\bigr) = \ln 2.$$

Add the two results:

$$2 - 2\ln 2 + \ln 2 = 2 - \ln 2.$$

The question states that the integral equals $$\alpha - \log_e 2$$, i.e.

$$\alpha - \ln 2 = 2 - \ln 2 \;\Longrightarrow\; \alpha = 2.$$

Finally, $$\alpha^{3} = 2^{3} = 8.$$

Hence the required value is $$8$$.