Question:
If $\frac{x}{y}+\frac{y}{x}=-1$ (where $x, y \neq 0$ ), then the value of $x^{3}-y^{3}$ is
(a) 1 -
(b) $-1$
(c) 0
(d) $\frac{1}{2}$
Solution:
(c)
Given, $\frac{x}{y}+\frac{y}{x}=-1$
$\Rightarrow$ $\frac{x^{2}+y^{2}}{x y}=-1$
$\Rightarrow$ $x^{2}+y^{2}=-x y$
$\Rightarrow$ $x^{2}+y^{2}+x y=0$ $\ldots($ i)
Now, $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$
[using identity, $\left.a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\right]$
$=(x-y) \times 0=0$ [from Eq. (i)]
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