Is the cube of Brownian motion a martingale?

Let $W$ be standard Brownian motion. Decide whether $W_t^3$ is a martingale, and if not, say how to fix it.

No. Ito gives $d(W^3)=3W^2\,dW+3W\,dt$, and the drift $3W_t\,dt$ is not zero; for instance $\mathbb{E}[W_t^3\mid\mathcal F_s]=W_s^3+3(t-s)W_s$. Subtracting the compensator gives the martingale $W_t^3-3tW_t$. In general $W^n$ is a martingale only for $n=0,1$.