3. (10 points) In a $3 \times 3$ grid (each cell is a $1 \times 1$ square), place two identical pieces, with at most one piece per cell. There are $\qquad$ different ways to place the pieces. (If two placements can be made to coincide by rotation, they are considered the same placement).
【Analysis】We can discuss the problem by dividing it into different cases. The four corner positions have the same value, the center square has one value, and the remaining four squares have the same value. Therefore, we can calculate the different placement methods in three cases.

【Solution】According to the analysis, we divide it into three cases:
(1) When a chess piece is placed in the center, i.e., at $E$, and the other chess piece is placed in any of the outer positions, excluding symmetry, there are 2 different placement methods, i.e., $A E$, $B E$;
(2) When neither of the two chess pieces is in the center $E$, and one of them is at a corner, there are 4 different placement methods, i.e., $A B$, $A F$, $A H$, $A D$;
(3) When both chess pieces are at the corners, there are 2 different placement methods, i.e., $A C$, $A I$;
(4) When both chess pieces are in the 4 squares other than the corners and the center, there are 2 different placement methods, i.e., $B D$, $B H$.
\begin{tabular}{|c|c|c|}
\hline$A$ & $B$ & $C$ \\
\hline$D$ & $E$ & $F$ \\
\hline$G$ & $H$ & $I$ \\
\hline
\end{tabular}

In summary, there are: $2+4+2+2=10$ different placement methods.