# Find the value

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:

$\left(x^{3}-y^{3}\right)=(x-y)\left(x^{2}+y^{2}+x y\right)$

$=(x-y) \times 0$

$=0$