and also laplace trans form of sin^2t
Part A)$\displaystyle \frac{1}{s^3-s} = \frac{1}{(s)(s^2-1)} $ This is the product of two Laplace transforms (Hint: Convolution Theorem is of interest here) :
$\displaystyle \mathcal{L}^{-1}\lbrace F(s)G(s) \rbrace = f(t)*g(t) = \int_{t'=0}^{t'=t} f(t')g(t'-t)dt'$
What are $\displaystyle F(s) $ and $\displaystyle G(s) $? What are the corresponding $\displaystyle f(t) $ and $\displaystyle g(t) $?
Part B) Laplace transform of $\displaystyle sin^2(t) $ Recall that $\displaystyle sin^2(t) = \frac{1}{2}-\frac{cos(2t)}{2} $ and we also have the definition of the Laplace transform: $\displaystyle \mathcal{L}\lbrace f(t) \rbrace = \int_{t=0}^{t\rightarrow \infty} e^{-st}f(t)dt $. What do you get when you substitute the above trig identity in the definition of the laplace transform (Hint: you should notice the answer will be a combination of two simpler Laplace transforms)
Another method:
$\displaystyle \dfrac{1}{s^3-s} = \dfrac{s^2-(s^2-1)}{s^3-s} = \dfrac{s^2}{s(s^2-1)} - \dfrac{s^2-1}{s(s^2-1)} = \dfrac{s}{s^2-1} - \dfrac{1}{s}$
Those two have easy inverse Laplace Transforms.