Question

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

Answer 1

Consider
$\left|\begin{array}{ccc}x+y+2z& x& y\\ z& y+z+2x& y\\ z& x& z+x+2y\end{array}\right|\stackrel{c_1^1=c_1+c_2+c_3}{\to }\left|\begin{array}{ccc}2(x+y+z)& x& y\\ 2(x+y+z)& y+z+2x& y\\ 2(x+y+z)& x& z+x+z+x+2y\end{array}\right|$
Taking $2(x+y+z)$ common from $C_1$
$=2(x+y+z)\left|\begin{array}{ccc}1& x& y\\ 0& x+y+z& 0\\ 0& 0& x+y+z\end{array}\right|=2(x+y+z{)}^3\left|\begin{array}{ccc}1& x& y\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|$ (Taking out $x+y+z$ from $R_2$ and $R_3$)
Expanding along $R_1$ we get,
$=2(x+y+z{)}^3[(1-0)-x\left(0\right)+y\left(0\right)]$
$=2(x+y+z{)}^3(RHS)$.