Solution:
$\begin{array}{l}
\left|\begin{array}{ccc}
(\mathrm{x}-2)^{2} & (\mathrm{x}-1)^{2} & \mathrm{x}^{2} \\
(\mathrm{x}-1)^{2} & \mathrm{x}^{2} & (\mathrm{x}+1)^{2} \\
\mathrm{x}^{2} & (\mathrm{x}+1)^{2} & (\mathrm{x}+2)^{2}
\end{array}\right| \\
=\left|\begin{array}{ccc}
\mathrm{x}^{2}-4 \mathrm{x}+4 & \mathrm{x}^{2}-2 \mathrm{x}+1 & \mathrm{x}^{2} \\
\mathrm{x}^{2}-2 \mathrm{x}+1 & \mathrm{x}^{2} & \mathrm{x}^{2}+2 \mathrm{x}+1 \\
\mathrm{x}^{2} & \mathrm{x}^{2}+2 \mathrm{x}+1 & \mathrm{x}^{2}+4 \mathrm{x}+4
\end{array}\right| \\
=\left|\begin{array}{ccc}
-2 \mathrm{x}+3 & -2 \mathrm{x}+1 & -2 \mathrm{x}-1 \\
-2 \mathrm{x}+1 & -2 \mathrm{x}-1 & -2 \mathrm{x}-3 \\
\mathrm{x}^{2} & \mathrm{x}^{2}+2 \mathrm{x}+1 & \mathrm{x}^{2}+4 \mathrm{x}+4
\end{array}\right|\left[\mathrm{R}_{1}^{\prime}=\mathrm{R}_{1}-\mathrm{R}_{2} \& \mathrm{R}_{2}^{\prime}=\mathrm{R}_{2}-\mathrm{R}_{3}\right]
\end{array}$
$=\left|\begin{array}{ccc}2 & 2 & 2 \\ -2 x+1 & -2 x-1 & -2 x-3 \\ x^{2} & x^{2}+2 x+1 & x^{2}+4 x+4\end{array}\right|\left[R_{1}^{\prime}=R_{1}-R_{2}\right]$
$=2\left|\begin{array}{ccc}1 & 1 & 1 \\ -2 x+1 & -2 x-1 & -2 x-3 \\ x^{2} & x^{2}+2 x+1 & x^{2}+4 x+4\end{array}\right|\left[R_{1}^{\prime}=R_{1} / 2\right]$
$=2\left|\begin{array}{ccc}1 & -2 x+1 & x^{2} \\ 1 & -2 x-1 & x^{2}+2 x+1 \\ 1 & -2 x-3 & x^{2}+4 x+4\end{array}\right|$ [transforming row and column]
$=2\left|\begin{array}{ccc}0 & 2 & -2 x-1 \\ 0 & 2 & -2 x-3 \\ 1 & -2 x-3 & x^{2}+4 x+4\end{array}\right|$ [ $\left.R_{1}^{\prime}=R_{1}-R_{2} \& R_{2}^{\prime}=R_{2}-R_{3}\right]$
$=2\left|\begin{array}{ccc}0 & 0 & 2 \\ 0 & 2 & -2 x-3 \\ 1 & -2 x-3 & x^{2}+4 x+4\end{array}\right|\left[R_{1}^{\prime}=R_{1}-R_{2}\right]$
$=2\{0+0+2(0-2)\}$ [expansion by first row]
$=-8$