Question

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

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

Answer 1

The correct option is C 0

As $\frac{x}{y}+\frac{y}{x}=-1$
$\frac{x^2+y^2}{xy}=-1$
$x^2+y^2=-xy$
$x^2+y^2+xy=0....\left(i\right)$
$(x^3-y^3)=(x-y).(x^2+y^2+xy)....\left(ii\right)$
From (i) & (ii)
$(x^3-y^3)=0$