3. Let $x^{2}-\frac{1}{x^{2}}$ and $x^{3}-\frac{1}{x^{3}}$ both be integers for the positive number $x(\quad$.
(A) does not exist
(B) there is only one
(C) there are only two
(D) there are infinitely many
3. (B).
$\because x^{3}-\frac{1}{x}$ is an integer,
and $\left(x^{3}-\frac{1}{x^{3}}\right)^{2}=x^{6}-2+\frac{1}{x^{6}}$,
$\therefore x^{6}+\frac{1}{x^{6}}$ is also an integer.
$\because x^{2}-\frac{1}{x^{2}}$ is an integer,
and $\left(x^{2}-\frac{1}{x^{2}}\right)^{3}=x^{6}-3 x^{2}+\frac{3}{x^{2}}-\frac{1}{x^{6}}$
$$
=\left(x^{6}-\frac{1}{x^{6}}\right)-3\left(x^{2}-\frac{1}{x^{2}}\right) \text {, }
$$
$\therefore x^{6}-\frac{1}{x^{6}}$ is also an integer.
Thus, $x^{6}+\frac{1}{x^{6}}$ and $x^{6}-\frac{1}{x^{6}}$ added and subtracted give $2 x^{6}$ and $\frac{2}{x^{6}}$ which are also integers, and their product is 4.

Therefore, $2 x^{6}$ is a divisor of 4, so it can only be $1, 2, 4$. Accordingly, $x^{6}=\frac{1}{2}, 1, 2$. When $x^{6}=\frac{1}{2}$ and 2, $x^{3}=\frac{\sqrt{2}}{2}$ and $\sqrt{2}$, $x^{3}-\frac{1}{x^{3}}$ is not an integer.

Thus, only when $x^{6}=1$, i.e., $x=1$, $x^{2}-\frac{1}{x^{2}}$ and $x^{3}-\frac{1}{x^{3}}$ are both integers.