Remember that row operations don't affect the determinant:
[tex]\begin{vmatrix}x+a&x&x\\x&x+a&x\\x&x&x+a\end{vmatrix}=\begin{vmatrix}a&a&0\\0&a&a\\-a&0&a\end{vmatrix}[/tex]
(that is, subtract row 2 from row 1; subtract row 3 from row 2; and subtract row 1 from row 3. This is not the only way to do it, of course.)
Factoring out [tex]a[/tex] gives
[tex]a^3\begin{vmatrix}1&1&0\\0&1&1\\-1&0&1\end{vmatrix}[/tex]
Notice that adding rows 1 and 3 gives the same numbers in row 2. In other words, the rows are not linearly independent, which means the matrix is singular, and this is so for any value of [tex]x[/tex].
