67 If $x$ is a root of the equation $\sqrt[3]{x+9}-\sqrt[3]{x-9}=3$, then $x^{2}$ lies between ( )
A. 55 and 56
B. 65 and 75
C. 75 and 85
D. 85 and 95
67 C. Cubing both sides of $(x+9)^{\frac{1}{3}}-(x-9)^{\frac{1}{3}}=3$ yields
$$
\begin{array}{c}
x+9-(x-9)-3(x+9)^{\frac{2}{3}}(x-9)^{\frac{1}{3}}+3(x+9)^{\frac{1}{3}}(x-9)^{\frac{2}{3}}=27, \\
9=3(x+9)^{\frac{1}{3}}(x-9)^{\frac{1}{3}}\left[(x-9)^{\frac{1}{3}}-(x+9)^{\frac{1}{3}}\right], \\
9=3(x+9)^{\frac{1}{3}} \cdot(x-9)^{\frac{1}{3}} \cdot(-3),
\end{array}
$$

Therefore,
$$
\left(x^{2}-81\right)^{\frac{1}{3}}=-1 \text {, }
$$

Thus,
$$
x^{2}=80 \text {. }
$$