MCQ
Write the correct answer in the following: If $\frac{\text{x}}{\text{y}}+\frac{\text{y}}{\text{x}}=-1 \ (\text{x},\text{y}\neq0),$ the value of $\text{x}^3-\text{y}^3$ is.
- A$1$
- B$-1$
- ✓$0$
- D$\frac{1}{2}$
✓
Answer
Correct option: C.
$0$
Given, $\frac{\text{x}}{\text{y}}+\frac{\text{y}}{\text{x}}=-1$
$\Rightarrow\frac{\text{x}^2+\text{y}^2}{\text{xy}}=-1$
$\Rightarrow\text{x}^2+\text{y}^2=-\text{xy}$
$\Rightarrow\text{x}^2+\text{y}^2+\text{xy}=0$
Now, $\text{x}^3-\text{y}^3=(\text{x}-\text{y})(\text{x}^2+\text{xy}+\text{y}^2) \ ...(\text{i})$
$[\text{a}^3-\text{b}^3=(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)]$
$=(\text{x}-\text{y})\times0=0 [$From Eq. $(i)]$
$\Rightarrow\frac{\text{x}^2+\text{y}^2}{\text{xy}}=-1$
$\Rightarrow\text{x}^2+\text{y}^2=-\text{xy}$
$\Rightarrow\text{x}^2+\text{y}^2+\text{xy}=0$
Now, $\text{x}^3-\text{y}^3=(\text{x}-\text{y})(\text{x}^2+\text{xy}+\text{y}^2) \ ...(\text{i})$
$[\text{a}^3-\text{b}^3=(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)]$
$=(\text{x}-\text{y})\times0=0 [$From Eq. $(i)]$
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