Question

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

• A. $1$
• B. $-1$
• C. $0$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $0$

Given, $\frac{x}{y}+\frac{y}{x}=-1(x,y\ne 0)$
$\Rightarrow \frac{x^2+y^2}{xy}=-1$
$\Rightarrow x^2+y^2+xy=0$ ...(i)
Now, $x^3-y^3=(x-y)(x^2+xy+y^2)$
$=(x-y)\times 0$ (using (i))
Therefore, $x^3-y^3=0$
Option C is correct.