Determinants and monic polynomials I wish to show that 
$$ \det \begin{pmatrix} 
x & a & a & a\\
a & x & a & a\\
a & a & x & a\\
a & a & a & x
\end{pmatrix}=(x-a)^3(x+3a).$$
Obviously, I could expand it out and try to notice some factors as I go, but that seems rather tedious. Is there a particularly fast way of demonstrating the above identity?

$$ \det \begin{pmatrix} 
x & a & a & a\\
a & x & a & a\\
a & a & x & a\\
a & a & a & x
\end{pmatrix}~\quad R_1\to R_1-R_2~\&~ R_2\to R_2-R_3 ~\&~ R_3\to R_3-R_4\\$$
$$==- \det \begin{pmatrix} 
x-a & a-x & 0 & 0\\
0 & x-a & a-x & 0\\
0 & 0 & x-a & a-x\\
a & a & a & x
\end{pmatrix}$$
$$=-(x-a)^3\det \begin{pmatrix} 
1 & -1 & 0 & 0\\
0 & 1 & -1 & 0\\
0 & 0 & 1 & -1\\
a & a & a & x
\end{pmatrix}$$
Now if you expand this then you'll get answer
