Question

If $\left|\begin{array}{ccc}{x}^k& x^{k+2}& x^{k+3}\\ y^k& y^{k+2}& y^{k+3}\\ z^k& z^{k+2}& z^{k+3}\end{array}\right|=(x-y)(y-z)(z-x)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$ then find the value of $-k$.

Answer 1

Given $\left|\begin{array}{ccc}{x}^k& x^{k+2}& x^{k+3}\\ y^k& y^{k+2}& y^{k+3}\\ z^k& z^{k+2}& z^{k+3}\end{array}\right|=(x-y)(y-z)(z-x)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$ ..... $\left(1\right)$
Consider, $\left|\begin{array}{ccc}{x}^k& x^{k+2}& x^{k+3}\\ y^k& y^{k+2}& y^{k+3}\\ z^k& z^{k+2}& z^{k+3}\end{array}\right|$
$=x^k{y}^k{z}^k\left|\begin{array}{ccc}1& x^2& x^3\\ 1& y^2& y^3\\ 1& z^2& z^3\end{array}\right|$
$R_1\to R_1-R_2,R_2\to R_2-R_3$

$=x^k{y}^k{z}^k\left|\begin{array}{ccc}0& (x-y)(x+y)& (x-y)(x^2+xy+y^2)\\ 0& (y-z)(y+z)& (y-z)(y^2+yz+z^2)\\ 1& z^2& z^3\end{array}\right|$
$=\left(xyz{)}^k\right(x-y\left)\right(y-z)\left|\begin{array}{ccc}0& x+y& x^2+xy+y^2\\ 0& y+z& y^2+yz+z^2\\ 1& z^2& z^3\end{array}\right|$
$R_1\to R_1-R_2$
$=\left(xyz{)}^k\right(x-y\left)\right(y-z)\left|\begin{array}{ccc}0& x-z& (x-z)(x+y+z)\\ 0& y+z& y^2+yz+z^2\\ 1& z^2& z^3\end{array}\right|$

$=\left(xyz{)}^k\right(x-y\left)\right(y-z\left)\right(z-x)\left|\begin{array}{ccc}0& -1& -(x+y+z)\\ 0& y+z& y^2+yz+z^2\\ 1& z^2& z^3\end{array}\right|$
$=\left(xyz{)}^k\right(x-y\left)\right(y-z\left)\right(z-x\left)\right(xy+yz+xz)$
$=\left(xyz{)}^{k+1}\right(x-y\left)\right(y-z\left)\right(z-x)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$
So, on comparing with $\left(1\right),$ we get
$k+1=0$
$\Rightarrow -k=1$