1.88 The values of $x^{2}, y^{2}$ that make the expression $\frac{x^{3}}{y}-\frac{y^{3}}{x}$ equal to $112 \sqrt{3}$ are
(A) $x^{2}=1+\sqrt{3}, y^{2}=2+\sqrt{3}$.
(B) $x^{2}=2+\sqrt{3}, y^{2}=2-\sqrt{3}$.
(C) $x^{2}=7+4 \sqrt{3}, y^{2}=7-4 \sqrt{3}$.
(D) $x^{2}=1+2 \sqrt{3}, y^{2}=2-\sqrt{3}$.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
[Solution] From $\frac{x^{3}}{y}-\frac{y^{3}}{x}=\frac{x^{4}-y^{4}}{x y}=\frac{\left(x^{2}-y^{2}\right)\left(x^{2}+y^{2}\right)}{x y}$, by verification, when $x^{2}=7+4 \sqrt{3}, \quad y^{2}=7-4 \sqrt{3}$, the original expression $=112 \sqrt{3}$.
Therefore, the answer is (C).