$f(t) = t^3$ .......... $f(0) = 0$
$f'(t) = 3t^2$ .......... $f'(0) = 0$
$f''(t) = 6t$ .......... $f''(0) = 0$
$f'''(t) = 6$
$\mathcal{L} \left\{ f'''(t) \right\} = s^3 \mathcal{L} \left\{ f(t) \right\} - s^2 f(0) - s \, f'(0) - f''(0)$
$\mathcal{L} (6) = s^3 \mathcal{L} (t^3) - s^2 f(0) - s \, f'(0) - f''(0)$
$6 \, \mathcal{L} (1) = s^3 \mathcal{L} (t^3) - s^2(0) - s(0) - (0)$
$6 \left( \dfrac{1}{s} \right) = s^3 \mathcal{L} (t^3)$
$s^3 \mathcal{L} (t^3) = \dfrac{6}{s}$
$\mathcal{L} (t^3) = \dfrac{6}{s^4}$ answer
