I am trying to prove that $E(W_t^3|\mathcal{F}_s) = W_s^3 + 3(t-s)W_s$ where $W_t$ is a standard Brownian motion and $\mathcal{F}_s$ is the usually filtration. So far, I am able to figure out most of the steps, but then I get stuck. Below is my attempt.
My attempt:
Observe that $W_t^3 = (W_s + W_t-W_s)^3$. Then \begin{align} E((W_s + W_t-W_s)^3 | \mathcal{F}_s) &= E(W_s^3 + 3W_s^2(W_t-W_s)+3W_s(W_t-W_s)^2+(W_t-W_s)^3|\mathcal{F}_s)\\ &= W_s^3 + 3W_s^2E(W_t-W_s |\mathcal{F}_s)+ 3W_sE((W_t-W_s)^2|\mathcal{F}_s) + E((W_t-W_s)^3|\mathcal{F}_s)\\ &= W_s^3 + 0\ + 3W_s(t-s)+E((W_t-W_s)^3|\mathcal{F}_s)\\ &=W_s^3 + 3W_s(t-s)+E((W_t-W_s)^3|\mathcal{F}_s). \end{align}
This is the point where I get stuck. I do not understand how to show that $E((W_t-W_s)^3|\mathcal{F}_s) = 0$.
For $t>s$, $W_t-W_s$ is independent of $\mathcal F_s$. Hence, $E((W_t-W_s)^{3}|\mathcal F_s)=E((W_t-W_s)^{3}$. Since $(W_t-W_s)^{3}$ has a symmetric distribution, its expectation is $0$.
[All odd moments of $N(0,\sigma^{2})$ are $0$].