Question

Prove that

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

Answer 1

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

$by{C}_1=C_1-C_2$

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

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

$by{C}_2=C_2-C_3$

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

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

$by{C}_1=C_1-C_2$

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

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

Expanding along $C_1$

$\Rightarrow \Delta =(x-y)(y-z)(x-z)\{0+1(xy+xz)-1(z^2-z^2-yz\left)\right\}$

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