3. Given $x, y>0, x^{2}+y^{2}=24$, $(\sqrt{x}+\sqrt{y})^{4}+(\sqrt{x}-\sqrt{y})^{4}=180$. Then $x y=(\quad)$.
(A) 8
(B) 9
(C) 10
(D) 11
3. D.

Notice,
$$
\begin{array}{l}
x^{2}+y^{2} \\
=(\sqrt{x}+\sqrt{y})^{4}-2(2 x \sqrt{x y}+3 x y+2 y \sqrt{x y}), \\
x^{2}+y^{2} \\
=(\sqrt{x}-\sqrt{y})^{4}+2(2 x \sqrt{x y}-3 x y+2 y \sqrt{x y}) .
\end{array}
$$
(1) + (2) gives
$$
\begin{array}{l}
2\left(x^{2}+y^{2}\right)=(\sqrt{x}+\sqrt{y})^{4}+(\sqrt{x}-\sqrt{y})^{4}-12 x y . \\
\text { Therefore } x y=\frac{180-24 \times 2}{12}=11 .
\end{array}
$$