$14 \cdot 20$ Find the real solutions of the equation $3 x^{3}-[x]=3$.
(30th International Mathematical Olympiad Candidate Problem, 1989)
[Solution] (1) If $x \leqslant -1$, then
$$
3 x^{3}-[x] \leqslant 3 x - x + 13 x - x \geqslant 4,
$$
(4) If $0 \leqslant x < 1$, then
$$
3 x^{3}-[x]=3 x^{3}<3 \text{. }
$$

In all the above cases, the equation
$$
3 x^{3}-[x]=3
$$

has no real solutions.
The last case.
(5) If $1 \leqslant x < 2$, then
$$
\begin{array}{c}
3 x^{3}-[x]=3 x^{3}-1=3 . \\
x=\sqrt[3]{\frac{4}{3}} .
\end{array}
$$

Therefore, the given equation has only one solution $x=\sqrt[3]{\frac{4}{3}}$.