To solve this, we start by evaluating the integral:

$ I = \int_0^{e^3} \left[ \frac{1}{e^{x-1}} \right] \, dx $

The greatest integer function $[\cdot]$ returns the largest integer less than or equal to the input value. Here's how we can approach the problem:

Determine the function inside the integral:

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

Identifying the intervals:

When $e^{1-x} \geq 2$, which simplifies to $x \leq 1 - \ln 2$, we have $\left[e^{1-x}\right] = 2$.

When $1 \leq e^{1-x} < 2$, simplifying gives $1 - \ln 2 < x \leq 1$, and thus $\left[e^{1-x}\right] = 1$.

When $0 \leq e^{1-x} < 1$, which holds for $x > 1$, thus $\left[e^{1-x}\right] = 0$ from $x = 1$ to $x = e^3$.

Evaluate the integral on these intervals:

$ \int_0^{1-\ln 2} 2 \, dx = 2(1-\ln 2) $

$ \int_{1-\ln 2}^1 1 \, dx = 1 - (1 - \ln 2) = \ln 2 $

$ \int_1^{e^3} 0 \, dx = 0 $

Combine these results:

$ I = 2(1-\ln 2) + \ln 2 + 0 = 2 - \ln 2 $

Thus, we are given that:

$ \alpha - \ln 2 = 2 - \ln 2 $

This implies that:

$ \alpha = 2 $

Therefore, $\alpha^3 = 2^3 = 8$.