= Solution
Summing the exact cubic increment identity gives
$$
B_t^3=\sum_j(\Delta B_j)^3
+3\sum_jB_j(\Delta B_j)^2
+3\sum_jB_j^2\Delta B_j.
$$
The first sum tends to zero in <L2 space>. Indeed, odd <moments> of centered <normal random variables> vanish, a centered <normal random variable> of <variance> $v$ has sixth <moment> $15v^3$, and <independence> of the <Brownian increments> gives
$$
\mathbb E\left|\sum_j(\Delta B_j)^3\right|^2
=15\sum_j(\Delta t_j)^3\leq15t|\pi|^2\longrightarrow0.
$$
The second sum converges in <L2 space> to $3\int_0^tB_sds$ by (i).
For the last sum, first note that $\mathbb E\int_0^tB_s^4ds=\int_0^t3s^2ds=t^3$, so $B_s^2$ is a <square-integrable stochastic integrand>. If $s=t_j+u$ and $D=B_s-B_j$, then $D$ is <independent> of $B_j$, with $D\sim N(0,u)$. Hence
$$
\mathbb E|B_s^2-B_j^2|^2
=\mathbb E|2B_jD+D^2|^2=4t_ju+3u^2.
$$
It follows that
$$
\mathbb E\int_0^t|B_s^2-B_{\eta(s)}^2|^2ds
\leq2t^2|\pi|+t|\pi|^2\longrightarrow0.
$$
The <Itô isometry> therefore makes $\sum_jB_j^2\Delta B_j$ converge in <L2 space> to $\int_0^tB_s^2dB_s$, with bounded truncations supplying the initial <simple predictable processes> if needed. Taking limits in the cubic identity proves
$$
\boxed{\int_0^tB_s^2\,dB_s=\frac13B_t^3-\int_0^tB_s\,ds.}
$$
The proof uses only increment algebra, <moments> of <normal random variables> and the definition of the <Itô integral>.
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