对于线性方程组
\begin{equation}
\left\{
\begin{array}{c}
x_0+a_1x_1+a_1^2x_2+\cdots+a_1^{n-1}x_{n-1}=b_1,\\
x_0+a_2x_1+a_2^2x_2+\cdots+a_2^{n-1}x_{n-1}=b_2,\\
\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\
x_0+a_nx_1+a_n^2x_2+\cdots+a_n^{n-1}x_{n-1}=b_n,
\end{array}\right.\tag{3.4.2}
\end{equation}
由于系数矩阵的行列式
\begin{equation*}
\det A=\begin{vmatrix}
1&a_1&a_1^2&\cdots&a_1^{n-1}\\
1&a_2&a_2^2&\cdots&a_2^{n-1}\\
\vdots&\vdots&\vdots&&\vdots\\
1&a_n&a_n^2&\cdots&a_n^{n-1}
\end{vmatrix}=\prod\limits_{1\leq k<l\leq n}(a_l-a_k)\neq 0,
\end{equation*}
\begin{equation*}
x_{j}=\frac{\det D_j}{\det A},\ j=0,1,\ldots,n-1,
\end{equation*}
其中
\begin{equation*}
\det D_j=\begin{vmatrix}
1&a_1&\cdots&a_1^{j-1}&b_1&a_1^{j+1}&\cdots&a_1^{n-1}\\
1&a_2&\cdots&a_2^{j-1}&b_2&a_2^{j+1}&\cdots&a_2^{n-1}\\
\vdots&\vdots&&\vdots&\vdots&\vdots&&\vdots\\
1&a_n&\cdots&a_n^{j-1}&b_n&a_n^{j+1}&\cdots&a_n^{n-1}
\end{vmatrix}.
\end{equation*}
因此存在唯一的次数小于 \(n\) 的多项式
\begin{equation*}
L(x)=c_0+c_1x+\cdots+c_{n-1}x^{n-1},
\end{equation*}
这里
\begin{equation*}
c_j=\frac{\det D_j}{\det A},\ j=0,1,\ldots,n-1,
\end{equation*}
使得对于任意的 \(i(1\leq i\leq n)\) ,都有 \(L(a_i)=b_i\)。下面证明
\begin{equation*}
L(x)=\sum\limits_{i=1}^n \left(b_i\cdot\prod\limits_{\begin{array}{c}1\leq j\leq n\\j\neq i\end{array}}\frac{x-a_j}{a_i-a_j}\right).
\end{equation*}
将 \(\det D_j\)按第 \(j+1\)列展开,有
\begin{equation*}
\det D_j=\sum\limits_{i=1}^n(-1)^{i+j+1}b_iM_{i,j+1},
\end{equation*}
其中
\begin{equation*}
M_{i,j+1}=\begin{vmatrix}
1&a_1&\cdots&a_1^{j-1}&a_1^{j+1}&\cdots&a_1^{n-1}\\
\vdots&\vdots&&\vdots&\vdots&&\vdots\\
1&a_{i-1}&\cdots&a_{i-1}^{j-1}&a_{i-1}^{j+1}&\cdots&a_{i-1}^{n-1}\\
1&a_{i+1}&\cdots&a_{i+1}^{j-1}&a_{i+1}^{j+1}&\cdots&a_{i+1}^{n-1}\\
\vdots&\vdots&&\vdots&\vdots&&\vdots\\
1&a_n&\cdots&a_n^{j-1}&a_n^{j+1}&\cdots&a_n^{n-1}
\end{vmatrix},
\end{equation*}
\begin{equation*}
\begin{array}{ccl}
c_j&=&\frac{\sum\limits_{i=1}^n(-1)^{i+j+1}b_i\left(\prod\limits_{\begin{array}{c}
1\leq k<l\leq n\\
k,l\neq i\end{array}}(a_l-a_k)\right)\cdot\left(\sum\limits_{\begin{array}{c}1\leq i_1< \cdots<i_{n-1-j}\leq n\\i_1,\ldots ,i_{n-1-j}\neq i\end{array}}a_{i_1}\cdots a_{i_{n-1-j}}\right)}{\prod\limits_{1\leq k<l\leq n}(a_l-a_k)}\\
&=&\sum\limits_{i=1}^n\left(\prod\limits_{\begin{array}{c} 1\leq k\leq n\\k\neq i\end{array}}\frac{b_i}{a_i-a_k}\right)\cdot\left((-1)^{n-j-1}\sum\limits_{\begin{array}{c}1\leq i_1< \cdots<i_{n-1-j}\leq n\\i_1,\ldots ,i_{n-1-j}\neq i\end{array}}a_{i_1}\cdots a_{i_{n-1-j}}\right).
\end{array}
\end{equation*}
由Viète定理可知
\begin{equation*}
(-1)^{n-j-1}\sum\limits_{\begin{array}{c}1\leq i_1< \cdots<i_{n-1-j}\leq n\\i_1,\ldots ,i_{n-1-j}\neq i\end{array}}a_{i_1}\cdots a_{i_{n-1-j}}
\end{equation*}
是多项式\(\prod\limits_{\begin{array}{c} 1\leq k\leq n\\k\neq i\end{array}}(x-a_k)\) 中 \(x^j\)的系数,因此 \(c_j\)是多项式
\begin{equation*}
\sum\limits_{i=1}^n \left(b_i\cdot\prod\limits_{\begin{array}{c} 1\leq k\leq n\\k\neq i\end{array}}\frac{x-a_k}{a_i-a_k}\right)
\end{equation*}
中 \(x^j\)的系数。从而
\begin{equation*}
L(x)=\sum\limits_{i=1}^n \left(b_i\cdot\prod\limits_{\begin{array}{c} 1\leq k\leq n\\k\neq i\end{array}}\frac{x-a_k}{a_i-a_k}\right).
\end{equation*}