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

• A. k = 2
• B. K = -1
• C. k = 0
• D. k = 1

Answer 1

The correct option is B K = -1

$\left|\begin{array}{ccc}{x}^k& x^{k+2}& x^{k+3}\\ y^k& y^{k+2}& y^{k+3}\\ z^k& z^{k+2}& z^{k+3}\end{array}\right|$
$=x^k.y^k.z^k(x-y)(y-z)(z-x)(xy+yz+zx)=\left(xyz{)}^{k+1}\right(x-y\left)\right(y-z\left)\right(z-x)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\therefore k+1=0\Rightarrow k=-1$