问题
已知，有范德蒙行列式 $D_{n}$ $=$$\left|\begin{array}{ccc} 1 & 1 & 1 \\ x_{1} & x_{2} & x_{3} \\ x_{1}^{2} & x_{2}^{2} & x_{3}^{2} \end{array}\right|$.
则，下面对该行列式的计算结果，正确的是哪个？
选项
[A]. $D$ $=$ $\left(x_{2}-x_{1}\right)$ $\cdot$ $\left(x_{3}-x_{1}\right)$[B]. $D$ $=$ $\left(x_{2}+x_{1}\right)$ $\cdot$ $\left(x_{3}+x_{1}\right)$ $\cdot$ $\left(x_{3}+x_{2}\right)$
[C]. $D$ $=$ $\left(x_{2}-x_{1}\right)$ $+$ $\left(x_{3}-x_{1}\right)$ $+$ $\left(x_{3}-x_{2}\right)$
[D]. $D$ $=$ $\left(x_{3} – x_{2} – x_{1}\right)$
$\left|\begin{array}{ccc} 1 & 1 & 1 \\ x_{1} & x_{2} & x_{3} \\ x_{1}^{2} & x_{2}^{2} & x_{3}^{2} \end{array}\right|$ $=$ $\left(x_{2}-x_{1}\right)$ $\cdot$ $\left(x_{3}-x_{1}\right)$ $\cdot$ $\left(x_{3}-x_{2}\right)$
范德蒙行列式的通用计算方式如下：
$\left|\begin{array}{ccccc} 1 & 1 & 1 & \cdots & 1 \\ x_{\textcolor{orange}{1}} & x_{\textcolor{orange}{2}} & x_{\textcolor{orange}{3}} & \cdots & x_{\textcolor{orange}{n}} \\ x_{\textcolor{orange}{1}}^{\textcolor{cyan}{2}} & x_{\textcolor{orange}{2}}^{\textcolor{cyan}{2}} & x_{\textcolor{orange}{3}}^{\textcolor{cyan}{2}} & \cdots & x_{\textcolor{orange}{n}}^{\textcolor{cyan}{2}} \\ \vdots & \vdots & \vdots & & \vdots \\ x_{\textcolor{orange}{1}}^{\textcolor{cyan}{n-1}} & x_{\textcolor{orange}{2}}^{\textcolor{cyan}{n-1}} & x_{\textcolor{orange}{3}}^{\textcolor{cyan}{n-1}} & \cdots & x_{\textcolor{orange}{n}}^{\textcolor{cyan}{n-1}} \end{array}\right|$ $=$
$\prod_{1 \leq j < i \leq n}$ $\left(x_{i} – x_{j} \right)$
总结：用右边的元素把左边的元素全减一遍并相乘。