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Let $x_1,x_2,\ldots,x_n$ be distinct real numbers. Take $p=0,1,2,\ldots$. I have checked using Mathematica that the following identity holds. \begin{equation} \sum\limits_{l=0}^{n-1} \frac{ \left|\begin{array}{cccccc} 1 & x_1 &\cdots& x_1^{l-1} & x_1^{l+p} & x_1^{l+1}& \cdots & x_1^{n-1}\\ 1 & x_2 &\cdots& x_2^{l-1} & x_2^{l+p} & x_2^{l+1}& \cdots &x_2^{n-1}\\ \vdots & \vdots & & \vdots & \vdots & \vdots&& \vdots\\ 1 & x_n &\cdots & x_n^{l-1} & x_n^{l+p} & x_n^{l+1}& \cdots &x_n^{n-1} \end{array} \right| }{ \left|\begin{array}{cccc} 1 & x_1 &\cdots& x_1^{n-1}\\ 1 & x_2 &\cdots& x_2^{n-1}\\ \vdots & \vdots & \ddots & \vdots \\ 1 & x_n &\cdots& x_n^{n-1} \end{array} \right| } = \sum\limits_{l=1}^n x_l^p \end{equation} How do I prove this identity?

Now, it would be a natural question to ask what happens if we insert an additional factor, say l, under the sum. In this case another identity holds: \begin{equation} \frac{1}{n} \sum\limits_{l=0}^{n-1} l \frac{\det\left(x_i^{j-1+p \delta_{j,l+1}}\right)}{\det\left(x_i^{j-1}\right)} = \frac{1}{n}\left[ (n-\frac{p+1}{2}) \sum\limits_{i=1}^n x_i^p + \frac{1}{2} \sum\limits_{q=1}^{p-1} \left(\sum\limits_{i=1}^n x_i^{p-q}\right) \left(\sum\limits_{i=1}^n x_i^q\right) \right] \end{equation} Here $p=1,2,3,\dots$. I have ``proven'' those identities by generating the matrices for particular values of $n$ and $p$ in Mathematica and then checking that both sides of the equation are the same.

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