1. **Problem Statement:** Find the Laplace Transform of the function $$f(t) = t^3 \cos t$$. 2. **Recall the Laplace Transform formula:** The Laplace Transform of $$t^n \cos(at)$$ is given by $$\mathcal{L}\{t^n \cos(at)\} = \frac{d^n}{ds^n} \left( \frac{s}{s^2 + a^2} \right) \cdot (-1)^n$$ where $$n$$ is a non-negative integer. 3. **Apply the formula for $$n=3$$ and $$a=1$$:** We want $$\mathcal{L}\{t^3 \cos t\} = (-1)^3 \frac{d^3}{ds^3} \left( \frac{s}{s^2 + 1} \right) = - \frac{d^3}{ds^3} \left( \frac{s}{s^2 + 1} \right)$$ 4. **Compute the first derivative:** Let $$F(s) = \frac{s}{s^2 + 1}$$. Using quotient rule: $$F'(s) = \frac{(1)(s^2 + 1) - s(2s)}{(s^2 + 1)^2} = \frac{s^2 + 1 - 2s^2}{(s^2 + 1)^2} = \frac{1 - s^2}{(s^2 + 1)^2}$$ 5. **Compute the second derivative:** $$F''(s) = \frac{d}{ds} \left( \frac{1 - s^2}{(s^2 + 1)^2} \right)$$ Using quotient rule again: Numerator derivative: $$-2s (s^2 + 1)^2 - (1 - s^2) \cdot 2 (s^2 + 1) \cdot 2s = -2s (s^2 + 1)^2 - 4s (1 - s^2)(s^2 + 1)$$ Denominator: $$(s^2 + 1)^4$$ Simplify numerator: $$-2s (s^2 + 1)^2 - 4s (1 - s^2)(s^2 + 1) = -2s (s^2 + 1) [ (s^2 + 1) + 2(1 - s^2) ]$$ Inside bracket: $$(s^2 + 1) + 2 - 2s^2 = s^2 + 1 + 2 - 2s^2 = 3 - s^2$$ So numerator: $$-2s (s^2 + 1)(3 - s^2)$$ Therefore, $$F''(s) = \frac{-2s (s^2 + 1)(3 - s^2)}{(s^2 + 1)^4} = \frac{-2s (3 - s^2)}{(s^2 + 1)^3}$$ 6. **Compute the third derivative:** $$F'''(s) = \frac{d}{ds} \left( \frac{-2s (3 - s^2)}{(s^2 + 1)^3} \right)$$ Rewrite numerator: $$-2s (3 - s^2) = -6s + 2s^3$$ Using quotient rule: Numerator derivative: $$(-6 + 6s^2)(s^2 + 1)^3 - (-6s + 2s^3) \cdot 3 (s^2 + 1)^2 \cdot 2s$$ Denominator: $$(s^2 + 1)^6$$ Simplify numerator stepwise: First term: $$(-6 + 6s^2)(s^2 + 1)^3$$ Second term: $$- (-6s + 2s^3) \cdot 6s (s^2 + 1)^2 = (6s - 2s^3) 6s (s^2 + 1)^2 = (36s^2 - 12 s^4)(s^2 + 1)^2$$ So numerator: $$(-6 + 6s^2)(s^2 + 1)^3 + (36 s^2 - 12 s^4)(s^2 + 1)^2$$ Factor out $$(s^2 + 1)^2$$: $$ (s^2 + 1)^2 [(-6 + 6s^2)(s^2 + 1) + 36 s^2 - 12 s^4]$$ Expand inside bracket: $$(-6 + 6s^2)(s^2 + 1) = -6 s^2 - 6 + 6 s^4 + 6 s^2 = 6 s^4 - 6$$ Add remaining terms: $$6 s^4 - 6 + 36 s^2 - 12 s^4 = (6 s^4 - 12 s^4) + 36 s^2 - 6 = -6 s^4 + 36 s^2 - 6$$ Therefore numerator: $$(s^2 + 1)^2 (-6 s^4 + 36 s^2 - 6)$$ 7. **Final expression for $$F'''(s)$$:** $$F'''(s) = \frac{(s^2 + 1)^2 (-6 s^4 + 36 s^2 - 6)}{(s^2 + 1)^6} = \frac{-6 s^4 + 36 s^2 - 6}{(s^2 + 1)^4}$$ 8. **Recall the Laplace Transform:** $$\mathcal{L}\{t^3 \cos t\} = - F'''(s) = - \frac{-6 s^4 + 36 s^2 - 6}{(s^2 + 1)^4} = \frac{6 s^4 - 36 s^2 + 6}{(s^2 + 1)^4}$$ **Final answer:** $$\boxed{\mathcal{L}\{t^3 \cos t\} = \frac{6 s^4 - 36 s^2 + 6}{(s^2 + 1)^4}}$$ This matches the given expression, confirming the correctness.