Question : If $\frac{1}{a}–\frac{1}{b}=\frac{1}{a–b}$, then the value of $a^{3}+b^{3}$ is:
Option 1: 0
Option 2: –1
Option 3: 1
Option 4: 2
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Correct Answer: 0
Solution : Given: $\frac{1}{a}–\frac{1}{b}=\frac{1}{a–b}$ ⇒ $\frac{b–a}{ab}=\frac{1}{a–b}$ ⇒ $(b–a)(a–b)=ab$ ⇒ $(a–b)^{2}=–ab$ ⇒ $a^{2}+b^{2}–2ab=–ab$ ⇒ $a^{2}+b^{2}=ab$ ⇒ $a^{2}+b^{2}-ab=0$ We know $a^3+b^3=(a+b)(a^2+b^2-ab)$ So, $a^3+b^3=0$ Hence, the correct answer is $0$.
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