= Solution
Put $M_t=B_t^3-3tB_t$. It is an <adapted process> and is integrable at every finite time, because <normal random variables> have finite absolute <moments> of every order. Fix $0\leq s<t$ and let $D=B_t-B_s$. Then $D$ is <independent> of $\mathcal F_s$, has <mean> zero, <variance> $t-s$, and third <moment> zero. Expanding the cube inside the <conditional expectation> gives
$$
\mathbb E[B_t^3\mid\mathcal F_s]
=B_s^3+3B_s\mathbb ED^2
=B_s^3+3(t-s)B_s.
$$
Also $\mathbb E[B_t\mid\mathcal F_s]=B_s$. Therefore
$$
\boxed{\mathbb E[M_t\mid\mathcal F_s]
=B_s^3+3(t-s)B_s-3tB_s
=B_s^3-3sB_s=M_s.}
$$
These are exactly the defining conditions for a <martingale>.
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