Question:
If $\frac{x}{y}+\frac{y}{x}=-1$, where $x \neq 0$ and $y \neq 0$, then the value of $\left(x^{3}-y^{3}\right)$ is
(a) 1
(b) $-1$
(c) 0
(d) $\frac{1}{2}$
Solution:
(c) 0
$\frac{x}{y}+\frac{y}{x}=-1$
$\Rightarrow \frac{x^{2}+y^{2}}{x y}=-1$
$\Rightarrow x^{2}+y^{2}=-x y$
$\Rightarrow x^{2}+y^{2}+x y=0$
Thus, we have:
$\left(x^{3}-y^{3}\right)=(x-y)\left(x^{2}+y^{2}+x y\right)$
$=(x-y) \times 0$
$=0$