4. For the equation $x^{3}-a x^{2}-2 a x+a^{2}-1=0$ with respect to $x$, if it has only one real root, then the range of values for $a$ is $\qquad$
4. $a<\frac{3}{4}$.

The original equation can be factored as
$$
[a-(x-1)]\left[a-\left(x^{2}+x+1\right)\right]=0 \text {. }
$$

Solving gives $a=x-1$ or $a=x^{2}+x+1$.
Thus, $x=a+1$ or $x^{2}+x+(1-a)=0$.
Since the original equation has only one real root, we have
$$
\Delta=1-4(1-a)<0 \text {. }
$$

Therefore, $a<\frac{3}{4}$.