Which one of the options given is the inverse Laplace transform of 1/ (𝑠³−𝑠 )? 𝑢(𝑡) denotes the unit-step function.

The correct option is:
$$\left(-1+\frac{1}{2}{e}^{-t}+\frac{1}{2}{e}^t\right)u\left(t\right)$$

Step-by-step Derivation:
We are given the Laplace domain function:
$$F\left(s\right)=\frac{1}{s^3-s}$$

First, factor the denominator:
$$s^3-s=s(s^2-1)=s(s-1)(s+1)$$

Thus, the expression becomes:
$$F\left(s\right)=\frac{1}{s(s-1)(s+1)}$$

Next, we perform a partial fraction decomposition:
$$\frac{1}{s(s-1)(s+1)}=\frac{A}{s}+\frac{B}{s-1}+\frac{C}{s+1}$$

We can determine the coefficients $A$, $B$, and $C$ using the cover-up method:
For $A$, multiply by $s$ and evaluate at $s=0$:
$$A={\left.\frac{1}{(s-1)(s+1)}\right|}_{s=0}=\frac{1}{(0-1)(0+1)}=-1$$

For $B$, multiply by $s-1$ and evaluate at $s=1$:
$$B={\left.\frac{1}{s(s+1)}\right|}_{s=1}=\frac{1}{1(1+1)}=\frac{1}{2}$$

For $C$, multiply by $s+1$ and evaluate at $s=-1$:
$$C={\left.\frac{1}{s(s-1)}\right|}_{s=-1}=\frac{1}{-1(-1-1)}=\frac{1}{2}$$

Substitute these values back into the partial fraction expansion:
$$F\left(s\right)=-\frac{1}{s}+\frac{1}{2(s-1)}+\frac{1}{2(s+1)}$$

Now, apply the inverse Laplace transform to each individual term:
- Using the standard transform pair ${\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}=u\left(t\right)$, we have:
$${\mathcal{L}}^{-1}\left\{-\frac{1}{s}\right\}=-u\left(t\right)$$
- Using the standard transform pair ${\mathcal{L}}^{-1}\left\{\frac{1}{s-a}\right\}=e^{at}u\left(t\right)$, we have:
$${\mathcal{L}}^{-1}\left\{\frac{1}{2(s-1)}\right\}=\frac{1}{2}{e}^{t}u\left(t\right)$$
and:
$${\mathcal{L}}^{-1}\left\{\frac{1}{2(s+1)}\right\}=\frac{1}{2}{e}^{-t}u\left(t\right)$$

Summing the individual inverse transforms yields:
$$f\left(t\right)=\left(-1+\frac{1}{2}{e}^{-t}+\frac{1}{2}{e}^t\right)u\left(t\right)$$