Example 2 Use $[a]$ to denote the greatest integer not exceeding the real number $a$, and $\{a\}=a-[a]$ to denote the fractional part of $a$. Solve the equation $\left[x^{3}\right]+\left[x^{2}\right]+[x]=\{x\}-1$.
(1991, Shanghai Junior High School Mathematics Competition)
Analysis: By definition, we have $0 \leqslant\{x\}<1$, which implies
$$
\begin{array}{l}
0 \leqslant\left[x^{3}\right]+\left[x^{2}\right]+[x]+1<1, \\
-1 \leqslant\left[x^{3}\right]+\left[x^{2}\right]+[x]<0 .
\end{array}
$$

Since $\left[x^{3}\right],\left[x^{2}\right],[x]$ are all integers, it follows that $\left[x^{3}\right]+ \left[x^{2}\right]+[x]=-1$. Therefore, $\{x\}=0$. Hence, $x$ is an integer, and we have $x^{3}+x^{2}+x+1=0$. Thus, the real solution to the equation is $x=-1$.