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Calculate the $n$-th order determinant:

$$\Delta= \begin{vmatrix} x&a&a&\ldots&a\\ a&x&a&\ldots&a\\ a&a&x&\ldots&a\\ \cdot&\cdot&\cdot&\ldots&\cdot\\ a&a&a&\ldots&x \end{vmatrix}$$

The answer is $\Delta=[x+a(n-1)](x-a)^{n-1}$.

If we add all the other columns to the first column, we get the first multiplicative factor of the answer, and are left with the following determinant:

$$\begin{vmatrix} 1&a&a&\ldots&a\\ 1&x&a&\ldots&a\\ 1&a&x&\ldots&a\\ \cdot&\cdot&\cdot&\ldots&\cdot\\ 1&a&a&\ldots&x \end{vmatrix}$$

How can we calculate this determinant to obtain the answer?