3. Given that $a, b$ are real numbers, if the quadratic function $f(x)=x^{2}+a x+b$ satisfies $f(f(0))=f(f(1))=0$, and $f(0) \neq f(1)$, then the value of $f(2)$ is $\qquad$ .
According to the problem, $f(0)=b, f(1)=1+a+b$ are the two distinct real roots of $x^{2}+a x+b=0$, so $\left\{\begin{array}{l}1+a+2 b=-a, \\ b(1+a+b)=b\end{array} \Rightarrow\left\{\begin{array}{l}a=-\frac{1}{2}, \\ b=0 .\end{array} \Rightarrow f(x)=x^{2}-\frac{1}{2} x\right.\right.$. Therefore, $f(2)=3$.