$14 \cdot 21 n$ is a positive integer, try to determine how many real numbers $x$, satisfy $1 \leqslant x<n$ and $x^{3}-\left[x^{3}\right]=(x-[x])^{3}$.
(Australian Mathematics Competition, 1992)
【Sol】Let $[x]=a, x=a+r, \quad 1 \leqslant a \leqslant n-1, \quad 0 \leqslant r<1$.
Then
$$
(a+r)^{3}-\left[(a+r)^{3}\right]=(a+r-[a+r])^{3} .
$$

So $a^{3}+3 a^{2} r+3 a r^{2}=\left[(a+r)^{3}\right]$.
Thus, $a^{3}+3 a^{2} r+3 a r^{2}$ is an integer. As $r$ increases from 0 to 1, $a^{3}+3 a^{2} r+3 a r^{2}$ increases from $a^{3}$ to $a^{3}+3 a^{2}+3 a$,
hence
$$
a^{3} \leqslant a^{3}+3 a^{2} r+3 a r^{2}<a^{3}+3 a^{2}+3 a=(a+1)^{3}-1 .
$$

Therefore, for $0 \leqslant r<1$, it takes $(a+1)^{3}-1-a^{3}$ integer values. Since $1 \leqslant a \leqslant n-1$, the number of real numbers $x$ that satisfy the condition is
$$
\begin{aligned}
& \sum_{a=1}^{n-1}\left[(a+1)^{3}-1-a^{3}\right] \\
= & \sum_{a=1}^{n-1}\left(3 a^{2}+3 a\right) \\
= & 3\left[\frac{(n-1) n(2 n-1)}{6}+\frac{n(n-1)}{2}\right] \\
= & n^{3}-n .
\end{aligned}
$$