3. Given that $a$ and $b$ 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 $\qquad$.
3. 3 .

It is known that $f(0)=b, f(1)=1+a+b$ are both roots of the equation $f(x)=0$.
$$
\begin{array}{l}
\text { Then } x^{2}+a x+b \equiv(x-b)(x-(1+a+b)) \\
\Rightarrow a=-1-a-2 b, b=b(1+a+b) \\
\Rightarrow a=-\frac{1}{2}, b=0 \\
\Rightarrow f(2)=3 .
\end{array}
$$