To find the inverse Laplace transform of $\frac{4}{s^3}$, we use the property:

$\mathcal{L}^{-1}\left(\frac{1}{s^{n+1}}\right) = \frac{t^n}{n!}, \quad n \geq 0$

Here, $\frac{4}{s^3}$ corresponds to $n = 2$. Applying this:

$\mathcal{L}^{-1}\left(\frac{4}{s^3}\right) = 4 \cdot \mathcal{L}^{-1}\left(\frac{1}{s^3}\right) = 4 \cdot \frac{t^2}{2!}$

Since $2! = 2$, we simplify:

$\mathcal{L}^{-1}\left(\frac{4}{s^3}\right) = 4 \cdot \frac{t^2}{2} = 2t^2$

Final Answer:

$\mathcal{L}^{-1}\left(\frac{4}{s^3}\right) = 2t^2$

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Would you like more details or have any questions?
Here are 5 related questions to explore further:

1. What is the Laplace transform of $t^n$?
2. How does the factorial term $n!$ appear in the inverse Laplace transform formula?
3. What is the significance of the Laplace transform in solving differential equations?
4. How would you find the Laplace inverse of $\frac{1}{s^n}$ for any general $n$?
5. Can you extend this to finding the inverse Laplace transform of $\frac{4}{(s+2)^3}$?

Tip: When working with the Laplace transform, always check the standard table of transforms for quick reference to frequently encountered expressions.