smc

By the Binomial Theorem, each term in the expansion is of the form $$\left(\genfrac{}{}{0ex}{}{1000}{k}\right){\left(x\sqrt[3]{2}\right)}^k{\left(y\sqrt{3}\right)}^{1000-k}=\left(\genfrac{}{}{0ex}{}{1000}{k}\right)2^{\frac{k}{3}}{3}^{\frac{1000-k}{2}}{x}^k{y}^{1000-k},$$ where $k\in \{0,1,2,\dots ,1000\}.$ This problem is equivalent to counting the values of $k$ such that both $\frac{k}{3}$ and $\frac{1000-k}{2}$ are integers. Note that $k$ must be a multiple of $3$ and a multiple of $2,$ so $k$ must be a multiple of $6.$ There are $\boxed{167}$ such values of $k:$ $$6(0),6(1),6(2),\dots ,6(166).$$