Proving that Brownian Motion Cubed is Martingale

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Hint: $E(\int_0^{t+s} W(u)du|\mathcal F_t)=\int_0^{t} W(u)du+E(\int_t^{t+s} [(W(u)-W(t))+W(t)]du|\mathcal F_t)$. The second term is just $sW(t)$ becasue $E(\int_t^{t+s} [(W(u)-W(t))]du=0$ by independence.