We know that for a submartingle $$A(t)$$, $$A(t)-\langle A\rangle_t$$ is a martingale where $$\langle A\rangle_t$$ is its quadratic variation.

For processes like $$W^3(t)$$ ($$W(t)$$ being standard Brownian Motion), It$$\hat{ \mathrm{o}}$$'s formula gives that $$d(W^3(t))=3W^2(t)dW(t)+3W(t)dt$$

Then we have $$W^3(t)-\int^t_03W(t)dt$$ is a martingale.

Can we conclude that $$\langle W^3(t)\rangle = \int^t_03W(s)ds$$ using the uniqueness of quadratic variation?

Furthermore, how can we calculate $$\int^t_03W(t)dt$$? It's not an It$$\hat{ \mathrm{o}}$$ integral.

Moreover, we can check that $$W^3(t)-3tW(t)$$ is a martingale. Does it imply $$\int^t_03W(s)ds = 3tW(t)$$?

The quadratic variation of a continuous local martingale $$X$$ (say) is the predictable increasing process $$V$$ such that $$X(t)^2-V(t)$$ is a local martingale. More generally, the quadratic variation of a continuous semimartingale coincides with the quadratic variation of its martingale part. In your example, this would be $$\langle W^3\rangle(t) =9\int_0^t W(s)^4\,ds$$.