$2 \cdot 129$ If $x$ is a number that satisfies $\sqrt[3]{x+9}-\sqrt[3]{x-9}=3$, then $x^{2}$ is between
(A) 55 and 65.
(B) 65 and 75.
(C) 75 and 85.
(D) 85 and 95.
(E) 95 and 105.
(14th American High School Mathematics Examination, 1963)
［Solution］Cubing both sides of the equation:
$$
\begin{aligned}
& (x+9)-3 \sqrt[3]{(x+9)^{2}(x-9)}+3 \sqrt[3]{(x+9)(x-9)^{2}} \\
& -(x-9) \\
= & 27
\end{aligned}
$$

This simplifies to $3 \sqrt[3]{(x+9)(x-9)}(\sqrt[3]{x+9}-\sqrt[3]{x-9})=-9$.
Assuming $\sqrt[3]{x+9}-\sqrt[3]{x-9}=3$, the equation further simplifies to
$$
\sqrt[3]{x^{2}-81}=-1 \text {, or } x^{2}=80 \text {. }
$$

Therefore, the answer is (C).