MCQ
If $\frac{a}{b}+\frac{b}{a}=-1$, then $a^3-b^3=$
- A1
- B-1
- C$\frac{1}{2}$
- ✓$0$
✓
Answer
Correct option: D.
$0$
(d)
We have,
$\frac{a}{b}+\frac{b}{a}=-1 \Rightarrow a^2+b^2+a b=0$
$\therefore \quad a^3-b^3=(a-b)\left(a^2+a b+b^2\right)=(a-b) \times 0=0$
We have,
$\frac{a}{b}+\frac{b}{a}=-1 \Rightarrow a^2+b^2+a b=0$
$\therefore \quad a^3-b^3=(a-b)\left(a^2+a b+b^2\right)=(a-b) \times 0=0$
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