I don't know how to exactly compute this determinant I've tried to compute this determinant by row transformations and column transformations, but it gives me a formula that doesn't work. The determinant is:
\begin{vmatrix}
x & a & b & c & d\\
a & x & b & c & d\\
a & b & x & c & d\\
a & b & c & x & d\\
a & b & c & d & x
\end{vmatrix}
I thought I could start doing row 5 - row 4, row 4 - row 3, row 3 - row 2 and row 2 - row 1 and then you get this determinant:
\begin{vmatrix}
x & a & b & c & d\\
a-x & x-a & 0 & 0 & 0\\
0 & b-x & x-b & 0 & 0\\
0 & 0 & c-x & x-c & 0\\
0 & 0 & 0 & d-x & x-d
\end{vmatrix}
Then I made column 1 - column 2, column 2 - column 3, column 3 - column 4, column 4 - column 5 and you get:
\begin{vmatrix}
x-a & a-b & b-c & c-d & d\\
0 & x-a & 0 & 0 & 0\\
0 & 0 & x-b & 0 & 0\\
0 & 0 & 0 & x-c & 0\\
0 & 0 & 0 & 0 & x-d
\end{vmatrix}
And, as it is triangular, you can multiply the diagonal elements, so you get that the determinant is:
$(x-a)^2(x-b)(x-c)(x-d)$
But this isn't correct and I don't know what to do, could someone please help me? I'd really appreciate.
 A: Notice that the sum of every row equals to $(a+b+c+d+x).$ You can take this approach:
$$\begin{vmatrix}
x & a & b & c & d\\
a & x & b & c & d\\
a & b & x & c & d\\
a & b & c & x & d\\
a & b & c & d & x
\end{vmatrix} \xrightarrow{\text{$C_1=C_1+C_2+C_3+C_4+C_5$}}  $$
$$ 
\begin{vmatrix}
(a+b+c+d+x) & a & b & c & d\\
(a+b+c+d+x) & x & b & c & d\\
(a+b+c+d+x) & b & x & c & d\\
(a+b+c+d+x) & b & c & x & d\\
(a+b+c+d+x) & b & c & d & x
\end{vmatrix} $$ $$= (a+b+c+d+x) \begin{vmatrix}
1 & a & b & c & d\\
1 & x & b & c & d\\
1 & b & x & c & d\\
1 & b & c & x & d\\
1 & b & c & d & x
\end{vmatrix}$$
Now, we can substract the first row from all of the other rows:
$$= (a+b+c+d+x) \begin{vmatrix}
1 & a & b & c & d\\
0 & x-a & 0 & 0 & 0\\
0 & b-a & x-b & 0 & 0\\
0 & b-a & c-b & x-c & 0\\
0 & b-a & c-b & d-c & x-d
\end{vmatrix} \xrightarrow{\text{Expand $C_1$}} $$
$$(a+b+c+d+x) \begin{vmatrix}
x-a & 0 & 0 & 0\\
b-a & x-b & 0 & 0\\
b-a & c-b & x-c & 0\\
b-a & c-b & d-c & x-d
\end{vmatrix} = $$ $$=(a+b+c+d+x)(x-a)(x-b)(x-c)(x-d)$$
A: Another approach: Consider it as a polynomial in $x$. Note that it will vanish when $x$ is any of $a,b,c,d$, since then we see a repeated row. It will also vanish when $x = -a-b-c-d$, since then all rows sum to $0$, and so we have the eigenvalue $0$ with eigenvector $(1,1,1,1,1)$.  Furthermore, we know that the leading coefficient is $1$.  Therefore the determinant is
$$(x-a)(x-b)(x-c)(x-d)(x-(-a -b-c-d)).$$
