14-22 Solve the equation $\left[x^{3}\right]+\left[x^{2}\right]+[x]=\{x\}-1$.
(Kiev Mathematical Olympiad, 1972)
[Solution] Since the left side of the given equation $\left[x^{3}\right]+\left[x^{2}\right]+[x]$ is an integer, the right side of the equation is also an integer. Also, because
$$
0 \leqslant\{x\}<1,
$$

it must be that
$$
\{x\}=0 .
$$

Thus, $x$ is an integer, and the equation simplifies to
$$
\begin{array}{c}
x^{3}+x^{2}+x=-1 \\
x^{3}+x^{2}+x+1=0 \\
(x+1)\left(x^{2}+1\right)=0 \\
x=-1 .
\end{array}
$$

Therefore, the equation has a unique solution: $x=-1$.