Correlation of Brownian motion with its own square

For a standard Brownian motion $B$ and fixed $t>0$, what is the correlation between $B_t$ and $B_t^2$?

$0$. $\operatorname{Cov}(B_t,B_t^2)=\mathbb{E}[B_t^3]-\mathbb{E}[B_t]\mathbb{E}[B_t^2]=0$, since odd moments of a centred normal vanish. They are uncorrelated even though $B_t^2$ is a function of $B_t$, a standard example that zero correlation does not imply independence.