Question

Prove $\left|\begin{array}{ccc}{x}^2& y^2& z^2\\ {(x+1)}^2& {(y+1)}^2& {(z+1)}^2\\ {(x-1)}^2& {(y-1)}^2& {(z-1)}^2\end{array}\right|=-4(x-y)(y-z)(x-z)$

Answer 1

$\left|\begin{array}{ccc}{x}^2& y^2& z^2\\ {(x+1)}^2& {(y+1)}^2& {(z+1)}^2\\ {(x-1)}^2& {(y-1)}^2& {(z-1)}^2\end{array}\right|=\left|\begin{array}{ccc}{x}^2-y^2& y^2-z^2& z^2\\ x^2-y^2+2(x-y)& y^2-z^2+2(y-z)& {(z+1)}^2\\ x^2-y^2-2(x-y)& y^2-z^2-2(y-z)& {(z-1)}^2\end{array}\right|=(x-y)(y-z)\left|\begin{array}{ccc}x+y& y+z& z^2\\ x+y+2& y+z+2& {(z+1)}^2\\ x+y-2& y+z-2& {(z-1)}^2\end{array}\right|=(x-y)(y-z)\left|\begin{array}{ccc}x-z& y+z& z^2\\ x-z& y+z+2& {(z+1)}^2\\ x-z& y+z-2& {(z-1)}^2\end{array}\right|=(x-y)(y-z)(x-z)\left|\begin{array}{ccc}1& y+z& z^2\\ 1& y+z+2& {(z+1)}^2\\ 1& y+z-2& {(z-1)}^2\end{array}\right|=(x-y)(y-z)(x-z)\left|\begin{array}{ccc}0& -2& 2z-1\\ 0& 4& 4z\\ 1& y+z-2& {(z-1)}^2\end{array}\right|=-4(x-y)(y-z)(x-z)$