At time t, $B_t \sim N(0,t)$. By symmetry, the $E[B_t^3]=0$.
Obviously, we know $E[B_t]=0$ from the question, but I am not sure how to reach $E[B_t^3]=0$ using symmetry of Brownian motion?
2026-04-29 09:40:05.1777455605
Expectation of cubic brownian motion: $E[B_t^3]$
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You have that $B_t \stackrel{d}{=} -B_t$ so that $$\mathbb{E}[B_t^3] = \mathbb{E}[-B_t^3]$$ which implies that $$\mathbb{E}[B_t^3] = 0.$$