1. **State the problem:** Find the Laplace Transform of the function $f(t) = t^3 \cos t$. 2. **Recall the formula:** The Laplace Transform of a function $f(t)$ is defined as $$\mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) \, dt$$ 3. **Use the known Laplace Transform formula for $t^n \cos(at)$:** $$\mathcal{L}\{t^n \cos(at)\} = \frac{(s^2 - a^2)(-1)^n n!}{(s^2 + a^2)^{n+1}}$$ for integer $n \geq 0$. 4. **Apply the formula for $n=3$ and $a=1$:** $$\mathcal{L}\{t^3 \cos t\} = \frac{(s^2 - 1)(-1)^3 3!}{(s^2 + 1)^4}$$ 5. **Calculate factorial and sign:** $3! = 6$ and $(-1)^3 = -1$. 6. **Substitute values:** $$\mathcal{L}\{t^3 \cos t\} = \frac{(s^2 - 1)(-1) \times 6}{(s^2 + 1)^4} = \frac{-6(s^2 - 1)}{(s^2 + 1)^4}$$ 7. **Simplify the numerator:** $$-6(s^2 - 1) = -6s^2 + 6$$ 8. **Final answer:** $$\boxed{\mathcal{L}\{t^3 \cos t\} = \frac{6 - 6s^2}{(s^2 + 1)^4}}$$ This is the Laplace Transform of $t^3 \cos t$.