Question

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

Answer 1

$\left|\begin{array}{ccc}x& x^2& yz\\ y& y^2& zx\\ z& z^2& xy\end{array}\right|$

Divide and multiply respectively 1st row by $x$ 2nd row by $y$ 3rd row by $z$

we get

$\frac{1}{xyz}\left|\begin{array}{ccc}{x}^2& x^3& xyz\\ y^2& y^3& xyz\\ z^2& z^3& xyz\end{array}\right|$

$\Delta =\frac{xyz}{xyz}\left|\begin{array}{ccc}{x}^2& x^3& 1\\ y^2& y^3& 1\\ z^2& z^3& 1\end{array}\right|$

$\Delta =\left|\begin{array}{ccc}{x}^2& x^3& 1\\ y^2& y^3& 1\\ z^2& z^3& 1\end{array}\right|$

$R_2\to R_2-R_1$

we get

$\Delta =\left|\begin{array}{ccc}{x}^2& x^3& 1\\ y^2-x^2& y^3-x^3& 0\\ z^2-x^2& z^3-x^3& 0\end{array}\right|$

Expand the determinant over 3rd column

We get

$\Delta =(y^2-x^2)(z^3-x^3)-(y^3-x^3)(z^2-x^2)$

$\Delta =(x-y)(y-z)(z-x)(xy+yz+zx)$ $\left[henceproved\right]$