Question

If $\frac{x}{y}+\frac{y}{x}=-1(x,y\ne 0)$, find the value of $x^3-y^3$

Answer 1

$\frac{x}{y}+\frac{y}{x}=-1$

$\Rightarrow \frac{x^2+y^2}{xy}=-1$

$\Rightarrow x^2+y^2=-1xy$

$\Rightarrow x^2+y^2+xy=0\dots \left(i\right)$

Using the identity, $x^3-y^3=(x-y)(x^2+xy+y^2)\dots \left(ii\right)$

on putting the value of equation (i) in equation (ii), we get

$\therefore x^3-y^3=0$