3. For any real value of $t$, the quadratic function
$$
y=\left(t^{2}+t+1\right) x^{2}-2(a+t)^{2} x+\left(t^{2}+3 a t+b\right)
$$

passes through the point $(1,0)$. Then $a=$ $\qquad$ ,$b=$ $\qquad$
3. 1,1 .

Let $t=0,1$, then we have
$$
\begin{array}{l} 
y=x^{2}-2 a^{2} x+b, \\
y=3 x^{2}-2(a+1)^{2} x+(1+3 a+b) .
\end{array}
$$

According to the problem, the graphs of functions (1) and (2) both pass through the point $(1,0)$, so,
$$
\begin{array}{l}
1-2 a^{2}+b=0, \\
3-2(a+1)^{2}+(1+3 a+b)=0 .
\end{array}
$$

Solving these equations, we get $a=1, b=1$.