3. Given the equation about $x$: $x^{3}+(1-a) x^{2}-2 a x+a^{2}$ $=0$ has only one real root. Then the range of the real number $a$ is $\qquad$.
3. $a<-\frac{1}{4}$.

The original equation is transformed into $(x-a)\left(x^{2}+x-a\right)=0$.
Thus, $x=a$ or $x^{2}+x-a=0$.
Since $x=a$ is the only real root of the equation, the equation $x^{2}+x-a=0$ has no real roots. Therefore, $\Delta=1+4a<0, a<-\frac{1}{4}$.