I desire to calculate $\mathbb E \{ B_s B_t ^2\} $, where $B$ is a standard brownian motion starting from zero. I want to be sure I am not making any mistake on both reasoning and result, even if I calculated it by three differents methods.
1° solution:
If $s\leq t$, then we have \begin{align}\mathbb E \{ B_s B_t ^2\} &= \mathbb E \{ B_s (B_t ^2-t)\} \\&= \mathbb E \{ B_s \mathbb E \{(B_t ^2-t) | \mathcal F_s\}\} \\&= \mathbb E \{ B_s (B_s ^2-s)\} \\&= \mathbb E \{ B_s ^3\} \\&=s^{3/2}\mathbb E \{ B_1 ^3\} =0\end{align}
If $t\leq s$, then we have \begin{align}\mathbb E \{ B_s B_t ^2\} &= \mathbb E \{ B_t ^2\mathbb E \{ B_s | \mathcal F_t\}\} \\&= \mathbb E \{ B_t ^3\} =0\end{align}
2° solution:
We know that $B^2_t = t +2 \int_0^t B_u ~dB_u$ and $B_s = \int_0^s ~dB_u$ then
\begin{align}\mathbb E \{ B_s B_t ^2\} &= 2 E \{\int_0^s ~dB_u\int_0^t B_u ~dB_u\} \\&=2 E \{\int_0^{t\wedge s} B_u ~du\} \\&=\int_0^{t\wedge s} 2 E \{B_u\} ~du =0\end{align}
3° solution:
It's trivial that $\mathbb E \{ B_s B_t ^2\} =0$ since $B_t^2 \geq 0$ and the brownian motion is gaussien process so it has symetric law.
I will be thankful for any feedback.