Question

Prove that $\begin{array}{ccc}(x+y{)}^2& zx& zy\\ zx& (z+y{)}^2& xy\\ zy& xy& (z+x{)}^2\end{array}$ = $2xyz(x+y+z{)}^3$

Answer 1

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

$=\frac{1}{xyz}\left|\begin{array}{ccc}z(x+y{)}^2& z^{2}x& z^{2}y\\ z{x}^2& x(z+y{)}^2& x^{2}y\\ z{y}^2& x{y}^2& y(z+x{)}^2\end{array}\right|$ [Multiplying first, second and third row by $z,x,y$ respectively]

$=\left|\begin{array}{ccc}(x+y{)}^2& z^2& z^2\\ x^2& (z+y{)}^2& x^2\\ y^2& y^2& (z+x{)}^2\end{array}\right|$ [Taking common $z,x,y$respectively from first, second and third row .]

[$C_2^{\prime }=C_2-C_1$ and $C_3^{\prime }=C_3-C_2$ gives the following]

$=\left|\begin{array}{ccc}(x+y{)}^2& z^2-(x+y{)}^2& 0\\ x^2& (z+y{)}^2-x^2& x^2-(z+y{)}^2\\ y^2& 0& (z+x{)}^2-y^2\end{array}\right|$

$=(x+y+z{)}^2\left|\begin{array}{ccc}(x+y{)}^2& z-(x+y)& 0\\ x^2& (z+y)-x& x-(z+y)\\ y^2& 0& (z+x)-y\end{array}\right|$

[$R_1^{\prime }=R_1-(R_2+R_3)$ gives the following]

$=(x+y+z{)}^2\left|\begin{array}{ccc}2\left(xy\right)& -2y& 2y-2x\\ x^2& (z+y)-x& x-(z+y)\\ y^2& 0& (z+x)-y\end{array}\right|$

[$C_3^{\prime }=C_3+C_2$ gives]

$=(x+y+z{)}^2\left|\begin{array}{ccc}2\left(xy\right)& -2y& -2x\\ x^2& (z+y)-x& 0\\ y^2& 0& (z+x)-y\end{array}\right|$

Now expanding with respect to the first row we get,

$=2(x+y+z{)}^2[xy(z+y-x)(z+x-y)+y{x}^2(z+x-y)+x{y}^2(z+y-x)]$

$=2xy(x+y+z{)}^2[(z+y-x)(z+x-y)+x(z+x-y)+y(z+y-x)]$

$=2xy(x+y+z{)}^2[(z+x-y)(z+y)+y(z+y-x)]$

$=2xy(x+y+z{)}^2[z(z+y+x)]$

$=2xyz(x+y+z{)}^3$