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?