$4$ by $4$ Determinant with variables I was given this $4$ by $4$ determinant
$$
\begin{vmatrix}
x & a & a & a \\
a & x & a & a \\
a & a & x & a \\
a & a & a & x 
\end{vmatrix} 
= 0 $$
Clearly one of the answers is $x=a,$ how do i find the other answer? 
I've tried splitting it into $4$  "$3$ by $3$" determinants but that didnt work well... 
 A: HINT
Add the $3$ last rows to the first one.
A: HINT: After subtracting, say, 4th row from first three you have enough zeros to compute the determinant easily.
A: Your matrix is the sum between a rank-$1$ matrix $M$ made by $a$s only and $(x-a)I$.
The spectrum of $M$ is $\{4a,0,0,0\}$ and the spectrum of $(x-a)I$ is $\{x-a,x-a,x-a,x-a\}$, so the spectrum of your matrix is $\{3a+x,(x-a),(x-a),(x-a)\}$ and the determinant (as the product of the eigenvalues) equals $(3a+x)(x-a)^3$. This is zero if $x=a$ or $x=-3a$.
A: Subtracting the 2nd row from the first, the 3rd from the second, the 4th from the third, tou obtain the determinant
$$\begin{vmatrix}
x-a&a-x&0&0\\0&x-a&a-x&0\\0&0&x-a&a-x\\ a&a&a&x
\end{vmatrix}=(x-a)^3\begin{vmatrix}
 1 & -1 & 0 & 0\\ 0 & 1 & -1 & 0\\ 0 & 0 & 1 & -1\\ a&a&a&x
\end{vmatrix}$$
Then, adding successively the 1st column to the first, the 2nd to the third, the 3rd to the fourth, we get an upper triangular determinant:
$$\begin{vmatrix}
 1 & -1 & 0 & 0\\ 0 & 1 & -1 & 0\\ 0 & 0 & 1 & -1\\ a&a&a&x
\end{vmatrix}=\begin{vmatrix}
 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ a&2a&3a&x+3a
\end{vmatrix}=x+3a.$$
A: Hint
There are many properties for solving determinant problems. One of them is that:

If you switch any column by a linear combination of any others, you don't change the determinant.

Using that, you can change the first column by the sum of all columns:
$$
\begin{vmatrix}
3a+x & a & a & a \\
3a+x & x & a & a \\
3a+x & a & x & a \\
3a+x & a & a & x 
\end{vmatrix} 
= 0 = (3a+x)\begin{vmatrix}
1 & a & a & a \\
1 & x & a & a \\
1 & a & x & a \\
1 & a & a & x 
\end{vmatrix}$$
Now you can use Laplace or Chió.
Can you finish?
