One. (20 points) Given the equation about $x$: $x^{3}-a x^{2}$ $-2 a x+a^{2}-1=0$ has only one real root. Find the range of real number $a$.


Consider the original equation as a quadratic equation in $a$, i.e.,
$$
a^{2}-\left(x^{2}+2 x\right) a+x^{3}-1=0 \text {. }
$$

Factoring, we get
$$
[a-(x-1)]\left[a-\left(x^{2}+x+1\right)\right]=0 .
$$

Thus, $x=a+1$ or $x^{2}+x+1-a=0$.
Since $x=a+1$ is not a root of equation (1), the original equation has exactly one real root when equation (1) has no real roots. Therefore,
$$
\Delta=1-4(1-a)<0 \text {. }
$$

Solving, we get $a<\frac{3}{4}$.