Question: 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?

Answer: 
No, I think you are right. The formula

$$\int\_0^t B\_s \, dB\_s = \frac{1}{2} (B\_t^2-t)$$

can be found in almost all books about SDE's. And you could also calculate the given moments using that

$$B\_t^2 -t = t \cdot \bigg( X\_t^2-1 \bigg)$$

where $X\_t := \frac{B\_t}{\sqrt{t}}$. $X\_t^2$ is $\chi\_1^2$-distributed and the third centered moment of a $\chi\_1^2$-random variable is equal to 8 (see [here](http://mathworld.wolfram.com/Chi-SquaredDistribution.html).) Hence

$$\mathbb{E}((B\_t^2-t)^3)=t^3 \cdot \mathbb{E}((X\_t^2-1)^3)=8t^3$$

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Remark: If you consider a process $X\_t := \int\_0^t \sigma(s) \, dB\_s$ where $\sigma$ is a deterministic(!) coefficient, then $X\_t$ is a centered normal random variable and in this case $\mathbb{E}(X\_t^3)=0$ holds.
For example, if $\mathbb{E}[X] = 0$, we cannot conclude that $\mathbb{E}[X^2] = 0$, etc. An example is $X = Z \sim \mathcal{N}(0, 1)$, where $\mathbb{E}[X] = 0$ but $\mathbb{E}[X^2] > 0$ since $X^2$ is never negative.