a,b,c are three real numbers where $a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a}$. Now $abc$ = ? Here will the answer be a number? I want to know whether it is possible to get a real number (not an algebraic expression)  as the product of $a$, $b$ and $c$.
I tried for a long time and this is what I got.
$$3abc = a^2 b + b^2 c + c^2 a$$
But it seems like there is a way to determine the value.
 A: Hint
\begin{align}
a-b&= \frac1c-\frac 1b \\ &=\frac{b-c}{bc}\\ &=\frac{c-a}{abc^2} \\& =\frac{a-b}{a^2b^2c^2}.
\end{align}
A: Substitute $$a+\frac{1}{b}=t,b+\frac{1}{c}=t,c+\frac{1}{a}=t$$ and solve this for $$a,b,c$$ this gives $$a=b=c=\frac{1}{2}(t-\sqrt{t^2-4})$$ or $$a=b=c=\frac{1}{2}(t+\sqrt{t^2-4})$$
A: We start with $$a + \frac 1b = b + \frac 1c = c + \frac 1a$$
This makes three equalities:
\begin{align}a + \frac 1b &= b + \frac 1c\tag{1}\\
b + \frac 1c &= c + \frac 1a\tag{2}\\
c + \frac 1a &= a + \frac 1b\tag{3}\end{align}
First consider equation $(1)$ $$a + \frac 1b = b + \frac 1c$$ which can be rearranged to give \begin{align}a + \frac 1b &= b + \frac 1c\\
a-b&=\frac 1c -\frac 1b\\
&=\frac{b-c}{bc}\end{align}
Similarly, from equation $(2)$, we can see that $$b-c=\frac{c-a}{ac}$$ and thus \begin{align}a-b&=\frac{b-c}{bc}\\
&=\frac{\frac{c-a}{ac}}{bc}\\
&=\frac{c-a}{abc^2}\end{align}
Finally, from equation $(3)$, we get $$c-a=\frac{a-b}{ab}$$ and thus \begin{align}a-b&=\frac{c-a}{abc^2}\\
&=\frac{\frac{a-b}{ab}}{abc^2}\\
&=\frac{a-b}{a^2b^2c^2}\end{align}
We can now rearrange this as follows (when $a\neq b$) \begin{align}a-b&=\frac{a-b}{a^2b^2c^2}\\
(a-b)(a^2b^2c^2)&=a-b\\
a^2b^2c^2&=\frac{a-b}{a-b}\\
a^2b^2c^2&=1\\
\sqrt{a^2b^2c^2}&=\sqrt1\\
\sqrt{a^2}\sqrt{b^2}\sqrt{c^2}&=\pm1\tag{$*$}\\
abc&=\pm1\end{align}
where $(*)$ comes from the fact that $\sqrt{a\cdot b} =\sqrt{a}\cdot\sqrt{b}$
A: If $a=b$ then $b=c$ and $a=b=c$, which says that $abc$ is not defined. 
Let $(a-b)(a-c)(b-c)\neq0.$
Thus, since 
$$a-b=\frac{1}{c}-\frac{1}{b}=\frac{b-c}{bc},$$
$$b-c=\frac{c-a}{ac}$$ and
$$c-a=\frac{a-b}{ab},$$ we obtain
$$(a-b)(b-c)(c-a)=\frac{(a-b)(b-c)(c-a)}{a^2b^2c^2},$$ which gives
$$abc=1$$ or $$abc=-1.$$
