Question

$\left|\begin{array}{ccc}x-y-z& 2x& 2x\\ 2y& y-z-z& xy\\ 2z& 2z& z-x-y\end{array}\right|$
Prove that: $\Delta =(x+y+z{)}^3$

Answer 1

$\left|\begin{array}{ccc}x-y-z& 2x& 2x\\ 2y& y-z-z& xy\\ 2z& 2z& z-x-y\end{array}\right|$
Applying $c_1-c_2\&c_2-c_3$
$\left|\begin{array}{ccc}-x-y-z& 0& 2x\\ y+z+x& -y-z-x& xy\\ 0x+y+z& & z-x-y\end{array}\right|$
Taking x+y+z common from $c_1\&c_2$
$(x+y+z{)}^2\left|\begin{array}{ccc}-1& 0& 2x\\ 1& -1& 2y\\ 0& 1& z-x-y\end{array}\right|$
Apply $R_1\to R_2+R_3$
$(x+y+z{)}^2\left|\begin{array}{ccc}0& 0& x+y+z\\ 1& -1& xy\\ 0& 1& z-x-y\end{array}\right|$
Expanding with $1^{st}row$
$\Delta =(x+y+z{)}^3$