To determine the inverse Laplace transform of $$\frac{1}{s^3 - s}$$, let's first rewrite the expression using partial fraction decomposition.

We have:

$$\frac{1}{s^3 - s} = \frac{1}{s(s^2 - 1)} = \frac{1}{s(s-1)(s+1)}$$

Using partial fraction decomposition, we can express it as:

$$\frac{1}{s(s-1)(s+1)} = \frac{A}{s} + \frac{B}{s-1} + \frac{C}{s+1}$$

To find the constants $ A $, $ B $, and $ C $, we equate the numerators:

$$1 = A(s-1)(s+1) + B s(s+1) + C s(s-1)$$

By solving for $ A $, $ B $, and $ C $, we get:

1. When $ s = 0 $:

$$1 = A(0-1)(0+1) + B(0)(0+1) + C(0)(0-1)$$

$$1 = -A \implies A = -1$$

2. When $ s = 1 $:

$$1 = A(1-1)(1+1) + B(1)(1+1) + C(1)(1-1)$$

$$1 = 2B \implies B = \frac{1}{2}$$

3. When $ s = -1 $:

$$1 = A(-1-1)(-1+1) + B(-1)(-1+1) + C(-1)(-1-1)$$

$$1 = -2C \implies C = -\frac{1}{2}$$

So, the partial fraction decomposition is:

$$\frac{1}{s(s-1)(s+1)} = -\frac{1}{s} + \frac{1}{2(s-1)} - \frac{1}{2(s+1)}$$

Next, we take the inverse Laplace transform of each term individually:

$$\mathcal{L}^{-1}\left(-\frac{1}{s}\right) = -1$$

$$\mathcal{L}^{-1}\left(\frac{1}{2(s-1)}\right) = \frac{1}{2}e^t$$

$$\mathcal{L}^{-1}\left(\frac{1}{2(s+1)}\right) = \frac{1}{2}e^{-t}$$

Combining these results, we get:

$$\mathcal{L}^{-1}\left(\frac{1}{s^3 - s}\right) = -1 + \frac{1}{2}e^t - \frac{1}{2}e^{-t}$$

Therefore, the inverse Laplace transform of $$\frac{1}{s^3 - s}$$ is:

$$\left(-1 + \frac{1}{2}e^t - \frac{1}{2}e^{-t}\right)u(t)$$

The correct option is Option A:

$$\left(-1 + \frac{1}{2}e^{-t} + \frac{1}{2}e^t\right)u(t)$$