Question

$\left|\begin{array}{ccc}x+y+2z& x& y\\ z& y+z+2x& y\\ z& x& z+x+2y\end{array}\right|=2(x+y+z{)}^3$

Answer 1

$LHS=\left|\begin{array}{ccc}x+y+2z& x& y\\ z& y+z+2x& y\\ z& x& z+x+2y\end{array}\right|=\left|\begin{array}{ccc}2(x+y+z)& x& y\\ 2(x+y+z)& y+z+2x& y\\ 2(x+y+z)& x& z+x+2y\end{array}\right|$
(using $C_1\to C_1+C_2+C_3$)

Take out $2(x+y+z)$ common from $C_1$, we get
$=2(x+y+z)\left|\begin{array}{ccc}1& x& y\\ 1& y+z+2x& y\\ 1& x& z+x+2y\end{array}\right|$

$=2(x+y+z)\left|\begin{array}{ccc}1& x& y\\ 0& y+z+x& 0\\ 0& 0& z+x+y\end{array}\right|$ (using $R_2\to R_2-R_1,R_3\to R_3-R_1$)

Take out $(x+y+z)$ common from $R_2$ and $R_3$, we get
$=2(x+y+z)(x+y+z)(x+y+z)\left|\begin{array}{ccc}1& x& y\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|$

Expanding along $R_3$, we get
$=2(x+y+z{)}^3[\left(1\right)(1-0)]=2(x+y+z{)}^3=RHS$.