W celu wyodrębnienia liczb całkowitych upraszczamy elementy zbioru A.

• $\frac{2-\sqrt{3}}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}\left(2-\sqrt{3}\right)}{3}=\frac{2\sqrt{3}-3}{3}\notin C$

• $\underline{\underline{6\sqrt{3}-3\sqrt{12}}}=6\sqrt{3}-3\sqrt{3\cdot 4}=6\sqrt{3}-3\cdot 2\sqrt{3}=6\sqrt{3}-6\sqrt{3}=0\in C$

• $\underline{\underline{\sqrt[3]{216}}}=6\in C$

• $\underline{\underline{3+\frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}-2\sqrt{6}}}=3+\frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}\cdot \frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}-2\sqrt{6}=$

$=3+\frac{{\left(3\sqrt{2}+2\sqrt{3}\right)}^2}{{\left(3\sqrt{2}\right)}^2-{\left(2\sqrt{3}\right)}^2}-2\sqrt{6}\stackrel{{\left(a+b\right)}^2=a^2+2ab+b^2}{=}3+\frac{18+12\sqrt{6}+12}{18-12}-2\sqrt{6}=$