Prove the following identities:

LHS

$=\left|\begin{array}{ccc}2 y & y-z-x & 2 y \\ 2 z & 2 z & z-x-y \\ x-y-z & 2 x & 2 x\end{array}\right|$

$=\left|\begin{array}{ccc}2 y+2 z+x-y-z & y-z-x+2 z+2 x & 2 y+z-x-y+2 x \\ 2 z & 2 z & z-x-y \\ x-y-z & 2 x & 2 x\end{array}\right|$ [Applying $\left.R_{1} \rightarrow R_{1}+R_{2}+R_{3}\right]$

$=\left|\begin{array}{ccc}x+y+z & x+y+z & x+y+z \\ 2 z & 2 z & z-x-y \\ x-y-z & 2 x & 2 x\end{array}\right|$

$=(x+y+z)\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 z & 2 z & z-x-y \\ x-y-z & 2 x & 2 x\end{array}\right| \quad$ [Taking $(x+y+z)$ common from $R_{1}$ ]

$=(x+y+z)\left|\begin{array}{ccc}0 & 1 & 1 \\ 0 & 2 z & z-x-y \\ -x-y-z & 2 x & 2 x\end{array}\right| \quad$ [Applying $C_{1} \rightarrow C_{1}-C_{2}$ ]

$=(x+y+z)^{2}\left|\begin{array}{ccc}0 & 1 & 1 \\ 0 & 2 z & z-x-y \\ -1 & 2 x & 2 x\end{array}\right| \quad\left[\right.$ Taking $(x+y+z)$ common from $\left.C_{1}\right]$

$=(x+y+z)^{2}[-1(z-x-y-2 z)] \quad$ [Expanding along first column]

$=(x+y+z)^{3}$

$=$ RHS

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