$$
A \coloneqq \begin{bmatrix}
x_{1} & x_{2} \\ x_{2} & x_{1}
\end{bmatrix},\ B \coloneqq \begin{bmatrix}
x_{3} & x_{4} \\ x_{4} & x_{3}
\end{bmatrix}$$
とおくと,p.55問2の結果を用いて
\begin{align}
\begin{vmatrix}
x_{1} & x_{2} & x_{3} & x_{4} \\
x_{2} & x_{1} & x_{4} & x_{3} \\
x_{3} & x_{4} & x_{1} & x_{2} \\
x_{4} & x_{3} & x_{2} & x_{1}
\end{vmatrix}
&= \begin{vmatrix}
A & B \\ B & A
\end{vmatrix} \\
&= \begin{vmatrix}
A + B
\end{vmatrix} \cdot \begin{vmatrix}
A-B
\end{vmatrix}\\
&= ((x_{1}+x_{3})^{2}-(x_{2}+x_{4})^{2}) \cdot ((x_{1}-x_{3})^{2}-(x_{2}-x_{4})^{2}) \\
&= (x_{1}+x_{2}+x_{3}+x_{4})(x_{1}-x_{2}+x_{3}-x_{4})(x_{1}+x_{2}-x_{3}-x_{4})(x_{1}-x_{2}-x_{3}+x_{4})
\end{align}
と計算できる.