Question

If $\left|\begin{array}{ccc}{x}^n& x^{n+2}& x^{n+4}\\ y^n& y^{n+2}& y^{n+4}\\ z^n& z^{n+2}& z^{n+4}\end{array}\right|=(\frac{1}{y^2}-\frac{1}{x^2})(\frac{1}{z^2}-\frac{1}{y^2})(\frac{1}{x^2}-\frac{1}{z^2})$, then $-n$ is ............

Answer 1

$\left|\begin{array}{ccc}{x}^n& x^{n+2}& x^{n+4}\\ y^n& y^{n+2}& y^{n+4}\\ z^n& z^{n+2}& z^{n+4}\end{array}\right|$
$\to x^n{y}^n{z}^n\left|\begin{array}{ccc}1& x^2& x^4\\ 1& y^2& y^4\\ 1& z^2& z^4\end{array}\right|$
applying $R_1\to R_1-R_3$ and $R_2\to R_2-R_3$ gives
$\to x^n{y}^n{z}^n(x^2-z^2)(y^2-z^2)\left|\begin{array}{ccc}0& 1& x^2+y^2\\ 0& 1& y^2+z^2\\ 1& z^2& z^4\end{array}\right|$
$=(x^2-y^2)(y^2-z^2)(z^2-x^2)x^n{y}^n{z}^n$
$\Rightarrow (x^2-y^2)(y^2-z^2)(z^2-x^2)x^n{y}^n{z}^n=(\frac{1}{y^2}-\frac{1}{x^2})(\frac{1}{z^2}-\frac{1}{y^2})(\frac{1}{x^2}-\frac{1}{z^2})$
$\Rightarrow (x^2-y^2)(y^2-z^2)(z^2-x^2)x^n{y}^n{z}^n=\frac{(x^2-y^2)(y^2-z^2)(z^2-x^2)}{x^4{y}^4{z}^4}$
$\Rightarrow n=-4$
$\therefore -n=4$