2. Given $x=\frac{1}{2+\sqrt{3}}, y=\frac{1}{2-\sqrt{3}}$. Then $x^{3}+12 x y+y^{3}=$ $\qquad$
2.64.

It is known that $x=2-\sqrt{3}, y=2+\sqrt{3}$. Therefore, $x+y=4$.
Then, $x^{3}+12 x y+y^{3}=x^{3}+y^{3}+3 x y(x+y)$ $=(x+y)^{3}=64$.