Question

If $\frac{x}{y}+\frac{y}{x}=-1$, where x ≠ 0 and y ≠ 0, then the value of (x³ − y³) is
(a) 1
(b) −1
(c) 0
(d) $\frac{1}{2}$

Answer 1

(c) 0

$$\begin{gathered} \frac{x}{y}+\frac{y}{x}=-1 \\ \Rightarrow \frac{x^2+y^2}{xy}=-1 \end{gathered}$$
$\Rightarrow x$² + y² = $-$xy
⇒ x² + y² + xy = 0

Thus, we have:
$\left(x^3-y^3\right)=\left(x-y\right)\left(x^2+y^2+xy\right)$
$$\begin{gathered} =\left(x-y\right)\times 0 \\ =0 \end{gathered}$$