1. **Problem:** Find the Laplace Transform of the function $f(t) = e^{-4t} \cos 3t + e^{-3t} t^3$. 2. **Formula and rules:** - The Laplace Transform of $e^{at} \cos(bt)$ is $\frac{s - a}{(s - a)^2 + b^2}$. - The Laplace Transform of $t^n e^{at}$ is $\frac{n!}{(s - a)^{n+1}}$. 3. **Apply the formula to each term:** - For $e^{-4t} \cos 3t$, here $a = -4$ and $b = 3$, so $\mathcal{L}\{e^{-4t} \cos 3t\} = \frac{s - (-4)}{(s + 4)^2 + 3^2} = \frac{s + 4}{(s + 4)^2 + 9}$. - For $e^{-3t} t^3$, here $a = -3$ and $n = 3$, so $\mathcal{L}\{t^3 e^{-3t}\} = \frac{3!}{(s + 3)^4} = \frac{6}{(s + 3)^4}$. 4. **Combine the results:** $\boxed{\mathcal{L}\{f(t)\} = \frac{s + 4}{(s + 4)^2 + 9} + \frac{6}{(s + 3)^4}}$. This is the Laplace Transform of the given function $f(t)$.