I want to calculate $\mathbb{E} \left[\left(\int_0^tB_s\text{d}B_s\right)^3\right]$ where $B_t$ is a standard Brownian motion. Using Ito's formula for $f:\mathbb{R}\rightarrow\mathbb{R}$ with $f(x)=x^2$ we can find that $\int_0^tB_s\text{d}B_s=\dfrac{B^2_t}{2}-\dfrac{t}{2}$. Then: $\mathbb{E} \left[\left(\int_0^tB_s\text{d}B_s\right)^3\right]=\frac{1}{8}\mathbb{E}\left[\left(B^2_t-t\right)^3\right]$ and $\mathbb{E}[(B^2_t-t)^3]=\mathbb{E}[B^6_t-3tB^4_t+3t^2B^2_t-t^3]=15t^3-9t^3+3t^3-t^3=8t^3$
However, I have seen in another question that $\mathbb{E} \left[\left(\int_0^tB_s\text{d}B_s\right)^3\right]=0$. Am I doing something wrong?