Question : If $a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}(a\neq b\neq c)$, then the value of $abc$ is:
Option 1: $\pm 1$
Option 2: $\pm 2$
Option 3: $0$
Option 4: $\pm \frac{1}{2}$
Correct Answer: $\pm 1$
Solution : $a+\frac{1}{b}=b+\frac{1}{c}$ ⇒ $(a-b) = \frac{(b-c)}{bc}$ ----------(i) $a+\frac{1}{b}=c+\frac{1}{a}$ ⇒ $(a-c) = \frac{(b-a)}{ba}$ ----------(ii) $b+\frac{1}{c}=c+\frac{1}{a}$ ⇒ $(b-c) = \frac{(c-a)}{ca}$ -----------(iii) Multiplying (i), (ii) and (iii), ⇒ $(a-b)(a-c)(b-c) = \frac{(b-c)(b-a)(c-a)}{bc×ba×ca}$ ⇒ $1 = \frac{1}{(abc)^2}$ ⇒ $abc = \pm 1$ Hence, the correct answer is $\pm 1$.
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