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This question already has an answer here:

the determinant is : $$\left|\begin{matrix} 1&-a_1&\cdots&-a^{n-1}_1\\ 1&a_2&\cdots&a^{n-1}_2\\ 1&a_3&\cdots&a^{n-1}_3\\ \vdots&\vdots&\ddots&\vdots\\ 1&a_n&\cdots&a_n^{n-1} \end{matrix}\right|$$

so i tried and got this :

$$\left|\begin{matrix} 1&0&0&\cdots&0\\ 1&a_2+a_1&a_2(a_2-a_1)&\cdots&a^{n-2}_2(a_2-a_1)\\ 1&a_3+a_1&a_3(a_3-a_1)&\cdots&a^{n-2}_3(a_3-a_1)\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&a_n+a_1&a_n(a_n-a_1)&\cdots&a_n^{n-2}(a_n-a_1) \end{matrix}\right|$$

and i dont know how to deal with the $a_k+a_1$ since i need the common factor of $a_k-a_1$. i would very much need a hint .

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marked as duplicate by user10354138, Community Jun 23 at 16:54

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Notice that $$\det\begin{pmatrix} 1&-a_1&\cdots&-a^{n-1}_1\\ 1&a_2&\cdots&a^{n-1}_2\\ 1&a_3&\cdots&a^{n-1}_3\\ \vdots&\vdots&\ddots&\vdots\\ 1&a_n&\cdots&a_n^{n-1} \end{pmatrix}=\det\begin{pmatrix} 2-1&-a_1&\cdots&-a^{n-1}_1\\ 1&a_2&\cdots&a^{n-1}_2\\ 1&a_3&\cdots&a^{n-1}_3\\ \vdots&\vdots&\ddots&\vdots\\ 1&a_n&\cdots&a_n^{n-1} \end{pmatrix}=\\=\det\begin{pmatrix} 2&-a_1&\cdots&-a^{n-1}_1\\ 0&a_2&\cdots&a^{n-1}_2\\ 0&a_3&\cdots&a^{n-1}_3\\ \vdots&\vdots&\ddots&\vdots\\ 0&a_n&\cdots&a_n^{n-1} \end{pmatrix}-\det\begin{pmatrix} 1&a_1&\cdots&a^{n-1}_1\\ 1&a_2&\cdots&a^{n-1}_2\\ 1&a_3&\cdots&a^{n-1}_3\\ \vdots&\vdots&\ddots&\vdots\\ 1&a_n&\cdots&a_n^{n-1} \end{pmatrix}$$

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