$$\frac{18}{3+\sqrt{3}}+3\sqrt{3}$$
$$= \frac{18(3-\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})} + 3\sqrt{3}$$
$$= \frac{18(3-\sqrt{3})}{9-3} + 3\sqrt{3} = \frac{18(3-\sqrt{3})}{6} + 3\sqrt{3} = 3(3-\sqrt{3}) + 3\sqrt{3} = 9-3\sqrt{3}+3\sqrt{3} = 9$$