<s>[INST] If $a+\frac1b=b+\frac1c=c+\frac1a$ for distinct $a$, $b$, $c$, how to find the value of $abc$? 
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.
 [/INST] I'm not sure how different this is, but here is my version
$a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a}
\quad$ (Note this implies $abc \ne 0$)
$a^2bc + ac = ab^2c + ab = abc^2 + bc$
$a(abc) + ac = b(abc) + ab = c(abc) + bc$
$\quad a(abc) + ac = b(abc) + ab \implies (a-b)abc=a(b-c)$
$$ a=b=c \ne 0 \; \text{ or } \; abc = \dfrac{a(b-c)}{a-b}$$
$\quad a(abc) + ac = c(abc) + bc \implies (a-c)abc = c(b-a)$
$$a=c=b \ne 0 \; \text{ or } \; abc = \dfrac{c(b-a)}{a-c}$$
$\quad b(abc) + ab = c(abc) + bc \implies (b-c)abc=b(c-a)$
$$b=c=a \ne 0 \; \text{ or } \; abc = \dfrac{b(c-a)}{b-c}$$
So, one solution is $\; a=b=c \ne 0$.
But if $a,b,$ and $c$ are distinct and non zero, then
$(abc)^3 =
 \dfrac{a(b-c)}{a-b} \cdot \dfrac{c(b-a)}{a-c} \cdot \dfrac{b(c-a)}{b-c}$
$(abc)^3 = abc$
$(abc)^2 = 1$
$abc = \pm 1$
$|abc| = 1$
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