Example 9 Given the equation in $x$, $x^{3}-a x^{2}$ $-2 a x+a^{2}-1=0$, which has only one real root. Then the range of the real number $a$ is $\qquad$ .
Solving: Rearrange the original equation into a quadratic equation in terms of $a$:
$$
\begin{array}{l}
a^{2}-x(x+2) a+x^{3}-1=0, \\
(a-x+1)\left(a-x^{2}-x-1\right)=0 . \\
\therefore x=a+1 \text { or } x^{2}+x+1-a=0 .
\end{array}
$$

According to the problem, the equation $x^{2}+x+1-a=0$ has no real roots.
Thus, $\Delta=1-4(1-a)<0$, which means $a<\frac{3}{4}$.