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 \( f(t) \).