To find the inverse Laplace transform of the function $\frac{1}{(s-4)^3}$, we can use the following formula:

$\mathcal{L}^{-1} \left\{ \frac{1}{(s-a)^n} \right\} = \frac{t^{n-1}}{(n-1)!} e^{at}$

Here, $a = 4$ and $n = 3$. Plugging in these values:

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

This corresponds to option d $\frac{t^2}{2} e^{4t}$.

So the correct answer is:

d. $\frac{t^2}{2} e^{4t}$

Would you like further details or have any questions? Here are some related questions:

1. What is the Laplace transform of $t^2 e^{4t}$?
2. How do you find the inverse Laplace transform of $\frac{1}{(s-a)^n}$ in general?
3. What is the inverse Laplace transform of $\frac{1}{(s-3)^4}$?
4. How can we apply the Laplace transform to solve differential equations?
5. What are the conditions for the existence of a Laplace transform?

Tip: Always remember that the inverse Laplace transform of a shifted function $\frac{1}{(s-a)^n}$ involves an exponential term $e^{at}$ multiplied by a polynomial in $t$.