3. The remainder of the polynomial $x^{20}+x^{10}+1$ divided by $(x-1)^{2}$ is . $\qquad$
3. $30 x-27$.

From the problem, we know that the degree of the remainder is no more than 1.
$$
\begin{array}{l}
\text { Let } x^{20}+x^{10}+1 \\
=(x-1)^{2} q(x)+a x+b,
\end{array}
$$

where $q(x)$ is the quotient, and $a, b$ are constants to be determined.
Let $x=1$, we get $3=a+b$.
Differentiating both sides of equation (1) yields
$$
\begin{array}{l}
20 x^{19}+10 x^{9} \\
=2(x-1) q(x)+(x-1)^{2} q^{\prime}(x)+a .
\end{array}
$$

Let $x=1$, we get $a=30$.
Thus, the remainder is $30 x-27$.