Problem 01 | Inverse Laplace Transform

$\mathcal{L}^{-1} \left[ \dfrac{8 - 3s + s^2}{s^3} \right] = \mathcal{L}^{-1} \left[ \dfrac{8}{s^3} - \dfrac{3s}{s^3} + \dfrac{s^2}{s^3} \right]$

$\mathcal{L}^{-1} \left[ \dfrac{8 - 3s + s^2}{s^3} \right] = 8\mathcal{L}^{-1} \left( \dfrac{1}{s^3} \right) - 3\mathcal{L}^{-1} \left( \dfrac{1}{s^2} \right) + \mathcal{L}^{-1} \left( \dfrac{1}{s} \right)$

$\mathcal{L}^{-1} \left[ \dfrac{8 - 3s + s^2}{s^3} \right] = 8\left[ \dfrac{t^{3 - 1}}{(3 - 1)!} \right] - 3\left[ \dfrac{t^{2 - 1}}{(2 - 1)!} \right] + 1$

$\mathcal{L}^{-1} \left[ \dfrac{8 - 3s + s^2}{s^3} \right] = 8\left( \dfrac{t^2}{2} \right) - 3\left( \dfrac{t}{1} \right) + 1$

$\mathcal{L}^{-1} \left[ \dfrac{8 - 3s + s^2}{s^3} \right] = 4t^2 - 3t + 1$           answer