Quadratic Variation of Brownian Motion Cubed

My question is simple. What is the quadratic variation of $W_t^3$ where $W_t$ is the standard one dimensional brownian motion.

My answer: I calculate $dX_tdX_t$ where $X_t = W_t^3$. Applying Ito's lemma to $X_t$, we have that

$dX_t = 3W_t^2dW_t + 3W_tdt$.

Hence $dX_tdX_t = 9W_t^4dt$ (using Ito calculus rules, $dW_tdW_t=dt$, $dW_tdt=0$ and $dtdt=0$) and therefore I conclude that the quadratic variation of $X_t$ is

$[X]_t = 9\int_{0}^{t} W_s^4 ds$.

However my friend says this is wrong. He says it is the process such that $W_t^6$ minus that process is a martingale and concludes (I don't know how) that the quadratic variation is $15\int_{0}^{t} W_s^4 ds$.

Who is correct?

What your friend was attempting was, seemingly to me at least, to use the fact that if $X_t$ is a local martingale, then $X_t^2-\left<X\right>_t$ is also a local martingale.
The tricky part is that, as you have shown, $${\rm d}\left(W_t^3\right)=3W_t^2{\rm d}W_t+3W_t{\rm d}t.$$ This implies that $$W_t^3-\int_0^t3W_s{\rm d}s=\int_0^t3W_s^2{\rm d}W_s$$ is a local martingale, while it is not the case for $W_t^3$. I think it could be that your friend mistook $W_t^3$ as a local martingale, and thus attempted to claim that $W_t^6-\left<W^3\right>_t$ is a local martingale.