Question : If the sum of $\frac{a}{b}$ and its reciprocal is 1 and $a\neq 0,b\neq 0$, then the value of $a^{3}+b^{3}$ is:
Option 1: 2
Option 2: –1
Option 3: 0
Option 4: 1
Correct Answer: 0
Solution : Given: $\frac{a}{b}$+$\frac{b}{a}= 1$ $⇒\frac{a^2+b^2}{ab}= 1$ $⇒a^2+b^2-ab= 0$---------(i) Now, $a^{3}+b^{3}$ $=(a+b)(a^2+b^2-ab)$ $= (a+b)×0$ $=0$ Hence, the correct answer is 0.
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Question : The numerical value of $\frac{(a–b)^{2}}{(b–c)(c–a)}+\frac{(b–c)^{2}}{(c–a)(a–b)}+\frac{(c–a)^{2}}{(a–b)(b–c)}$ is: $(a\neq b\neq c)$
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