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The trick is to start with the highest power, rewrite it as something you know (a third order moment) and then work backwards on the remaining terms. By that I mean you can complete the cube as follows: $$E[W_t^3 - 3tW_t|\mathcal{F}_s] = E[(W_t-W_s)^3 - C -3tW_t|\mathcal{F}_s]$$ where you'll need to find $C$ such that the equality holds (i.e. $C=W_s^3 + ... 1 You can use that$f(t,W_t)\in C^2$is Martingale iff:$$\partial_t f+\frac{1}{2}\partial_{WW}f= 0$$ We get:$$\partial_t f=-3W_t$$$$\partial_{WW}f=6W_t$$ Finally: $$-3W_t+3W_t= 0$$ q.e.d. The proof of theorem follows by writing out$f(t,W_t)\$ via Ito formula. Proof of theorem: