By using properties of determinants,show that:
$\left|\begin{array}{ccc}x+y+2z & x & y \\ z & y+z+2x & y \\ z & x & z+x+2y\end{array}\right|=2(x+y+z)^{3}$

(N/A) Let $\Delta = \left|\begin{array}{ccc}x+y+2z & x & y \\ z & y+z+2x & y \\ z & x & z+x+2y\end{array}\right|$.
Applying the column operation $C_{1} \rightarrow C_{1} + C_{2} + C_{3}$,we get:
$\Delta = \left|\begin{array}{ccc}2(x+y+z) & x & y \\ 2(x+y+z) & y+z+2x & y \\ 2(x+y+z) & x & z+x+2y\end{array}\right|$
Taking $2(x+y+z)$ common from $C_{1}$:
$\Delta = 2(x+y+z) \left|\begin{array}{ccc}1 & x & y \\ 1 & y+z+2x & y \\ 1 & x & z+x+2y\end{array}\right|$
Applying row operations $R_{2} \rightarrow R_{2} - R_{1}$ and $R_{3} \rightarrow R_{3} - R_{1}$:
$\Delta = 2(x+y+z) \left|\begin{array}{ccc}1 & x & y \\ 0 & x+y+z & 0 \\ 0 & 0 & x+y+z\end{array}\right|$
Taking $(x+y+z)$ common from $R_{2}$ and $R_{3}$:
$\Delta = 2(x+y+z)(x+y+z)(x+y+z) \left|\begin{array}{ccc}1 & x & y \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$
$\Delta = 2(x+y+z)^{3} \times (1(1-0)) = 2(x+y+z)^{3}$.
Hence,the result is proved.