TITLE: If $a+\frac1b=b+\frac1c=c+\frac1a$ for distinct $a$, $b$, $c$, how to find the value of $abc$?
QUESTION [10 upvotes]: If $a, b, c$ be distinct reals such that $$a + \frac1b = b + \frac1c = c + \frac1a$$ how do I find the value of $abc$?

The answer says $1$, but I am not sure how to derive it.

REPLY [2 votes]: let $a+\frac{1}{b} = b+\frac{1}{c} = c+\frac{1}{a} = k$
$ab + 1 = bk, bc + 1 = ck, ca + 1 = ak \Rightarrow ab=bk-1, bc=ck-1, ca=ak-1$
$abc + c = bck = (ck-1)k = ck^2-k \Rightarrow abc + k=c(k^2-1)$
$abc + a = ack = (ak-1)k = ak^2-k \Rightarrow abc+k =a(k^2-1)$
$abc+b=abk=(bk-1)k=bk^2-k \Rightarrow abc+k=b(k^2-1)$
$c(k^2-1)=a(k^2-1)=b(k^2-1)=abc + k$
$a\neq b \neq c \neq 0 \Rightarrow k^2-1=0 \Rightarrow k^2=1 \Rightarrow k= \pm1$
$abc+k=c(k^2-1)=0 \Rightarrow abc=-k=\pm1$