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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