MCQ
If $\frac{\text{x}}{\text{y}}+\frac{\text{y}}{\text{x}}=-1(\text{x},\text{y}\not=0),$ the value of $x^3 - y^3$ is:
- ✓$0$
- B$2$
- C$-1$
- D$1$
✓
Answer
Correct option: A.
$0$
Given: $\frac{\text{x}}{\text{y}}+\frac{\text{y}}{\text{x}}=-1$
$\Rightarrow\frac{\text{x}^2+\text{y}^2}{\text{xy}}=-1$
$\Rightarrow\text{x}^2+\text{y}^2=-\text{xy}\ ...(\text{i})$
Now, $\text{x}^3-\text{y}^3=(\text{x}^2+\text{y}^2+\text{xy})$
$\Rightarrow\text{x}^3-\text{y}^3=(\text{x}-\text{y})(\text{-xy}+\text{xy})\ \ \ [\text{from eq. (i)}]$
$\Rightarrow\text{x}^3-\text{y}^3=(\text{x}-\text{y})(0)$
$\Rightarrow(\text{x}^3-\text{y}^3)=0$
$\Rightarrow\frac{\text{x}^2+\text{y}^2}{\text{xy}}=-1$
$\Rightarrow\text{x}^2+\text{y}^2=-\text{xy}\ ...(\text{i})$
Now, $\text{x}^3-\text{y}^3=(\text{x}^2+\text{y}^2+\text{xy})$
$\Rightarrow\text{x}^3-\text{y}^3=(\text{x}-\text{y})(\text{-xy}+\text{xy})\ \ \ [\text{from eq. (i)}]$
$\Rightarrow\text{x}^3-\text{y}^3=(\text{x}-\text{y})(0)$
$\Rightarrow(\text{x}^3-\text{y}^3)=0$
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