$\mathcal{L} \left\{ t^n f(t) \right\} = (-1)^n \dfrac{d^n}{ds^n} F(s)$
$\mathcal{L} (\cos 3t) = \dfrac{s}{s^2 + 3^2}$
$\mathcal{L} (\cos 3t) = \dfrac{s}{s^2 + 9}$
$\mathcal{L} (t^2 \cos 3t) = (-1)^2 \dfrac{d^2}{ds^2} \left[ \dfrac{s}{s^2 + 9} \right]$
$\mathcal{L} (t^2 \cos 3t) = \dfrac{d}{ds} \left[ \dfrac{(s^2 + 9)(1) - s(2s)}{(s^2 + 9)^2} \right]$
$\mathcal{L} (t^2 \cos 3t) = \dfrac{d}{ds} \left[ \dfrac{s^2 + 9 - 2s^2}{(s^2 + 9)^2} \right]$
$\mathcal{L} (t^2 \cos 3t) = \dfrac{d}{ds} \left[ \dfrac{9 - s^2}{(s^2 + 9)^2} \right]$
$\mathcal{L} (t^2 \cos 3t) = \dfrac{(s^2 + 9)^2(-2s) - (9 - s^2)[ \, 2(s^2 + 9)(2s) \, ]}{[ \, (s^2 + 9)^2 \, ]^2}$
$\mathcal{L} (t^2 \cos 3t) = \dfrac{-2s(s^2 + 9)^2 - 4s(9 - s^2)(s^2 + 9)}{(s^2 + 9)^4}$
$\mathcal{L} (t^2 \cos 3t) = \dfrac{-2s(s^2 + 9) \, [ \, (s^2 + 9) + 2(9 - s^2) \, ]}{(s^2 + 9)^4}$
$\mathcal{L} (t^2 \cos 3t) = \dfrac{-2s \, [ \, s^2 + 9 + 18 - 2s^2 \, ]}{(s^2 + 9)^3}$
$\mathcal{L} (t^2 \cos 3t) = \dfrac{-2s(27 - s^2)}{(s^2 + 9)^3}$
$\mathcal{L} (t^2 \cos 3t) = \dfrac{2s(s^2 - 27)}{(s^2 + 9)^3}$ answer
