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Compute the determinant: $$ \det\begin{bmatrix} 1 & x_{1} & x_{1}^{2}& \dots & x_{1}^{n-2} & (x_{2}+ x_{3}+ \dots + x_{n})^{n-1} \\ 1 & x_{2} & x_{2}^{2}& \dots & x_{2}^{n-2} & (x_{1}+ x_{3}+ \dots + x_{n})^{n-1} \\ \vdots & \vdots &\vdots &\dots & \vdots& \vdots \\ 1 & x_{n} & x_{n}^{2}& \dots & x_{n}^{n-2} & (x_{1}+ x_{2}+ \dots + x_{n-1})^{n-1} \\ \end{bmatrix}.$$

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up vote 7 down vote accepted

Put $$ A(x_1,\ldots,x_n)=\begin{bmatrix} 1 & x_{1} & x_{1}^{2}& \dots & x_{1}^{n-2} & (x_{2}+ x_{3}+ \dots + x_{n})^{n-1} \\ 1 & x_{2} & x_{2}^{2}& \dots & x_{2}^{n-2} & (x_{1}+ x_{3}+ \dots + x_{n})^{n-1} \\ \vdots & \vdots &\vdots &\dots & \vdots& \vdots \\ 1 & x_{n} & x_{n}^{2}& \dots & x_{n}^{n-2} & (x_{1}+ x_{2}+ \dots + x_{n-1})^{n-1} \\ \end{bmatrix}.$$ Then we have \begin{align*} \det A(x_1,\ldots,x_n)&= \begin{vmatrix} 1 & x_{1} & x_{1}^{2}& \dots & x_{1}^{n-2} & (x_{2}+ x_{3}+ \dots +\ x_{n})^{n-1} \\ 1 & x_{2} & x_{2}^{2}& \dots & x_{2}^{n-2} & (x_{1}+ x_{3}+ \dots +\ x_{n})^{n-1} \\ \vdots & \vdots &\vdots&&\vdots& \vdots \\ 1 & x_{n} & x_{n}^{2}& \dots & x_{n}^{n-2} & (x_{1}+ x_{2}+ \dots +\ x_{n-1})^{n-1} \end{vmatrix}\\ &=\begin{vmatrix} 1 & x_{1} & x_{1}^{2}& \dots & x_{1}^{n-2} & \sum_{k=0}^{n-1}\binom {n-1}k(-1)^kx_1^k(x_1+\ldots + x_n)^{n-k-1} \\ 1 & x_{2} & x_{2}^{2}& \dots & x_{2}^{n-2} &\sum_{k=0}^{n-1}\binom {n-1}k(-1)^kx_2^k(x_1+\ldots +x_n)^{n-k-1} \\ \vdots & \vdots &\vdots&&\vdots& \vdots \\ 1 & x_{n} & x_{n}^{2}& \dots & x_{n}^{n-2} & \sum_{k=0}^{n-1}\binom {n-1}k(-1)^kx_n^k(x_1+\ldots + x_n)^{n-k-1} \end{vmatrix}\\ &=\sum_{k=0}^{n-1}\binom{n-1}k(-1)^k(x_1+\ldots +x_n)^{n-k-1}\begin{vmatrix} 1 & x_{1} & x_{1}^{2}& \dots & x_{1}^{n-2} & x_1^k \\ 1 & x_{2} & x_{2}^{2}& \dots & x_{2}^{n-2} &x_2^k \\ \vdots & \vdots&\vdots &&\vdots& \vdots \\ 1 & x_{n} & x_{n}^{2}& \dots & x_{n}^{n-2} &x_n^k \end{vmatrix}\\ &=(-1)^{n-1}\begin{vmatrix} 1 & x_{1} & x_{1}^{2}& \dots & x_{1}^{n-2} & x_1^{n-1} \\ 1 & x_{2} & x_{2}^{2}& \dots & x_{2}^{n-2} &x_2^{n-1} \\ \vdots & \vdots &\vdots&&\vdots& \vdots \\ 1 & x_{n} & x_{n}^{2}& \dots & x_{n}^{n-2} &x_n^{n-1} \end{vmatrix} \end{align*} hence
$$\det A(x_1,\ldots,x_n)=(-1)^{n-1}\prod_{1\leq i<j\leq n}(x_j-x_i).$$

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This is ingenious! – Paul Dec 11 '11 at 11:40

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