Making the Cube of Brownian Motion a Martingale

Let $$B_t$$ be a standard one dimensional Brownian
Motion. Find the function $$F:\mathbf{R}^5 \rightarrow \mathbf R$$ so that
\begin{align*}
B_t^3 – F\Big(t,B_t,B_t^2,\int_0^t B_s ds, \int_0^t B_s^2 ds\Big)
\end{align*}
is a Martingale.

Hint: It might be useful to introduce the processes
$X_t=B_t^2\qquad Y_t=\int_0^t B_s ds \qquad Z_t=\int_0^t B_s^2 ds$