Expected value of the Ito integral $\mathbb{E}\left[\int_0^T 4W^4_t \,\text{d}W_t\right]$

Question

I am confused on how to calculate the following $$\mathbb{E}\left[\int_0^T 4W^4_t \,\text{d}W_t\right]$$

Notes:

1. $W=(W_t)_{t\geq 0}$ is a Brownian motion with the filtration $\mathbb{F}=(\mathcal{F}_t)_{t\geq 0}$.
2. From 1., the integrand is of course adapted to $\mathbb{F}$.
3. This integral is called Ito integral.

From the solution manual I have, it says since integrals w.r.t $W_t$ are martingales, so the value is $0$. How to see this fact?