stochastic processes - Show that $E[B_t|mathscr{F}_s] = B_s$ for $B_t = W_t^3 - 3 t W_t$

Given prob space $(\Omega, \mathscr{F}, P)$ and a Wiener process $(W_t)_{t \geq 0}$, define filtration $\mathscr{F}_t = \sigma(W_u : u \leq t)$

Let $(B_t)_{t \geq 0}$ where $B_t = W_t^3 - 3tW_t$. Show that $E[B_t|\mathscr{F}_s] = B_s$ whenever $s < t$.

I think this all comes down to manipulation since there are martingales somewhere

My attempt:

Splitting up into $E[W_t^3|\mathscr{F}_s] - 3E[tW_t|\mathscr{F}_s]$ doesn't do anything since those guys aren't martingales? So, I tried splitting it up into:

$E[W_t(W_t^2 - 3t)|\mathscr{F}_s]$

$= E[W_t(W_t^2 - t -2 t)|\mathscr{F}_s]$

$= E[W_t(W_t^2 - t) -2 tW_t)|\mathscr{F}_s]$

$= E[W_t(W_t^2 - t)|\mathscr{F}_s] -2E[ tW_t|\mathscr{F}_s]$

$W_t$ is not $\mathscr{F}_s$-measurable, so we can't take that out...

$tW_{1/t}$ is Brownian and thus a martingale, but I don't know about $tW_t$...

$cW_{t/c^2}$ is Brownian and thus a martingale, but I don't think we can set c = t...

Help please?

Answer

\begin{align*} E\Big(W_t^3-3tW_t \mid \mathcal{F}_s\Big) &= E\Big((W_t-W_s+W_s)^3-3t(W_t-W_s+W_s) \mid \mathcal{F}_s\Big) \\ &=E\Big((W_t-W_s)^3+W_s^3+3(W_t-W_s)^2W_s + 3 (W_t-W_s)W_s^2\\ &\qquad \qquad -3t(W_t-W_s)-3tW_s \mid \mathcal{F}_s\Big) \\ &=E\Big((W_t-W_s)^3\Big) + W_s^3+3W_sE\Big((W_t-W_s)^2\Big)\\ &\qquad \qquad + 3W_s^2 E(W_t-W_s)-3tE(W_t-W_s)-3tW_s\\ &=W_s^3+3W_s (t-s)-3tW_s\\ &=W_s^3 -3sW_s, \end{align*} by noting that \begin{align*} E\Big((W_t-W_s)^3\Big) = E(W_t-W_s) =0, \end{align*} and \begin{align*} E\Big((W_t-W_s)^2\Big) = t-s. \end{align*}