Compute: $E\left[(B_s-B_t)^3\mid\mathcal{F}_t\right]$, $s>t$.
$B_t$ is standard 1D Brownian motion, and $\mathcal{F}_t=\sigma(B_t)$.
Here is my attempt:
$$ \begin{aligned} E\left[(B_s-B_t)^3\mid\mathcal{F}_t\right]&=E\left[(B_{s-t}-B_0)^3\mid\mathcal{F}_0\right] \\ &=E\left[(B_{s-t})^3\mid\mathcal{F}_0\right] \quad \text{ (since $B_0=0$)}\\ &=0 \quad \text{ (since $B_{s-t}\sim N(0,s-t)$)} \end{aligned} $$
Here are some other things I have deduced (are they ok?):
1) $E\left[(B_s-B_t)\right]=0 \quad$ ($B_s-B_t$ is $N(0,s-t)$)
2) $E\left[(B_s-B_t)\mid\mathcal{F}_t\right]=
E\left[B_s\mid\mathcal{F}_t\right]-B_t=0 \quad$ ($B_t$ martingale property)
3) $E\left[(B_s-B_t)^2\right]=s-t\quad$ (standard $B_t$ property)
4) $E\left[(B_s-B_t)^2\mid\mathcal{F}_t\right]=E\left[(B_{s-t})^2\mid\mathcal{F}_0\right]=s-t \quad$ (etc...)