jmc

Let $a=\sqrt{2}$ and $b=\sqrt[3]{3}.$ By the Binomial Theorem, $$(a+b{)}^{12}=(\genfrac{}{}{0ex}{}{12}{0})a^{12}+(\genfrac{}{}{0ex}{}{12}{1})a^{11}b+(\genfrac{}{}{0ex}{}{12}{2})a^{10}{b}^2+\cdots +(\genfrac{}{}{0ex}{}{12}{12})b^{12}.$$The term $a^k{b}^{12-k}$ is rational if and only if $k$ is divisible by 2, and $12-k$ is divisible by 3. Then $k$ must be divisible by 3, so $k$ must be a multiple of 6. Therefore, the sum of the rational terms is $$\left(\genfrac{}{}{0ex}{}{12}{0}\right)a^{12}+\left(\genfrac{}{}{0ex}{}{12}{6}\right)a^6{b}^6+\left(\genfrac{}{}{0ex}{}{12}{12}\right)b^{12}=\boxed{66673}.$$