What is $ 3E[\int_t^T (B_s^2+(T-s))dB_s|\mathcal{F}_t]=E[B_T^3-B_t^3]=? $ I try to compute
$$
M_t=E[B_T^3|\mathcal{F}_t]=E[3\int_0^T (B_s^2+(T-s))dB_s|\mathcal{F}_t]
$$
where using Ito fomula and integrate by part we have $B^3_T=3\int_0^T (B_s^2+(T-s))dB_s$.
$$
M_t=3E[\int_0^t (B_s^2+(T-s))dB_s+\int_t^T (B_s^2+(T-s))dB_s|\mathcal{F}_t]=3\int_0^t (B_s^2+(T-s))dB_s+3E[\int_t^T (B_s^2+(T-s))dB_s|\mathcal{F}_t]
$$
and I am stuck in
$$
3E[\int_t^T (B_s^2+(T-s))dB_s|\mathcal{F}_t]=E[B_T^3-B_t^3]=?
$$
Also, $E[B^3_t]=0?$
 A: I am not sure if that equation is true. From
\begin{align}
dB^2_t&=2B_t\,dB_t+dt\\
d\langle B^2,B\rangle_t&=2B_t\,d\langle B,B\rangle_t=2B_t\,dt
\end{align}
we get
\begin{align}
dB^3_t&=B_t\,dB^2_t+B^2_t\,dB_t+d\langle B^2,B\rangle_t\\
&=2B^2_t\,dB_t+B_t\,dt+B^2_t\,dB_t+2B_t\,dt\\
&=3B^2_t\,dB_t+3B_t\,dt\,.
\end{align}
Therefore,
\begin{align}\tag{1}
B_T^3-B_t^3&=\int_t^T3B^2_s\,dB_s+\int_t^T3B_s\,ds\,.
\end{align}
The term $M_T=\int_0^T3B^2_t\,dB_t$ is a martingale because it is an Ito integral of BM. Then from
$$
M_t=\mathbb E[M_T|{\cal F}_t]
$$
we find $\mathbb E[M_T-M_t|{\cal F}_t]=0\,.$ Now observe
$$
M_T-M_t=\int_t^T3B^2_s\,dB_s\,.
$$
We have shown that the conditional expectation of this is zero.
Applying this to (1) implies
\begin{align}\tag{2}
\boxed{\quad\mathbb E\Big[B^3_T-B^3_t\,\Big|\,{\cal F}_t\Big]=\mathbb E\Bigg[\int_t^T3B_s^2\,ds\,\Bigg|\,{\cal F}_t\Bigg]\,.\quad}
\end{align}
Regarding your many other questions in the comments, let me answer with a few hints:

*

*Because $M_T$ is a martingale and $M_0=0$ what does this mean for
$\mathbb E[M_T]=\mathbb E[\int_0^T3B^2_s\,dB_s]\,?$


*From the last equality (2) we get $\mathbb E[B^3_T-B^3_t]=\mathbb E[B^3_T-B^3_t\,|\,{\cal F}_t]$ but not that this is zero. Hint: is the integral in the RHS of (2) a $dB_s$-integral or a $ds$-integral?


*You get an expression for $\mathbb E[B^3_T|{\cal F}_t]$ by setting $t=0$ in (1) and using again that $M_T$ is a martingale.


*Why is your equation $B^3_t=\int_0^t3B_s^2\,ds+\int_0^t3B_s\,ds$ correct? Hint: what do you have to choose for $t$ and $T$ in (1)?


*$\mathbb E[B_T^3|{\cal F}_t]=B_t^3$ is wrong. Hint: can you deduce from (1) that $B_T^3$ is not a martingale?
