Question

if a+1/b=1=b+1/c then abc+1 is

Answer 1

Complete question is given below.

$\textsf{Find \ abc+1 \ if \ a+\dfrac{1}{b}=b+\dfrac{1}{c}=c+\dfrac{1}{a}=1. }$

$\cline{1-}$

$a+\dfrac{1}{b}=b+\dfrac{1}{c}\ \ \ \Rightarrow\ \ \ a-b=\dfrac{1}{c}-\dfrac{1}{b}=\dfrac{b-c}{bc}\ \ \ \ \ (1)\\ \\ \\ b+\dfrac{1}{c}=c+\dfrac{1}{a}\ \ \ \Rightarrow\ \ \ b-c=\dfrac{1}{a}-\dfrac{1}{c}=\dfrac{c-a}{ca}\ \ \ \ \ (2)\\ \\ \\ c+\dfrac{1}{a}=a+\dfrac{1}{b}\ \ \ \Rightarrow\ \ \ c-a=\dfrac{1}{b}-\dfrac{1}{a}=\dfrac{a-b}{ab}\ \ \ \ \ (3)$

Now,

$\begin{aligned}&(1) \times (2) \times (3)\\ \\ \Longrightarrow\ \ &(a-b)(b-c)(c-a)=\frac{(b-c)(c-a)(a-b)}{bc\cdot ca\cdot ab}\\ \\ \Longrightarrow\ \ &(a-b)(b-c)(c-a)=\frac{(a-b)(b-c)(c-a)}{(abc)^2}\\ \\ \Longrightarrow\ \ &(abc)^2=\frac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}\\ \\ \Longrightarrow\ \ &(abc)^2=1\\ \\ \Longrightarrow\ \ &abc=\pm 1\\ \\ \Longrightarrow\ \ &abc+1=1\pm 1\end{aligned}$

Hence the answer is 0 or 2.