Question

Using properties of determinants, prove that
$\omega \left|\begin{array}{ccc}x& y& z\\ x^2& y^2& z^2\\ y+z& z+x& x+y\end{array}\right|=(x-y)(y-z)(z-x)(x+y+z)$

Answer 1

Let $\Delta =\omega \left|\begin{array}{ccc}x& y& z\\ x^2& y^2& z^2\\ y+z& x+z& x+y\end{array}\right|$

Applying $R_3\to R_3+R_1$, we get

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

$=(x+y+z)\omega \left|\begin{array}{ccc}x& y& z\\ x^2& y^2& z^2\\ 1& 1& 1\end{array}\right|$

Applying $C_1\to C_1-C_2$ and $C_2\to C_2-C_3$, we get

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

$=(x+y+z)(x-y)(y-z)\omega \left|\begin{array}{ccc}1& 1& z\\ x+y& y+z& z^2\\ 0& 0& 1\end{array}\right|$

Expanding along $R_3$, we get

$=(x+y+z)(x-y)(y-z)(y+z-x-y)$

$=(x-y)(y-z)(z-x)(x+y+z)$