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.
 A: Below I explicitly highlight the symmetry implicit in the answers of Aryabhata and Andre.  Let $\rm\,\sigma\,$ be the permutation  $\rm\,(a\ b\ c)\,$ acting on polynomials in $\rm\,a,b,c\,$ by cyclically permuting the variables, i.e. $\rm\,\sigma\ f(a,b,c) = f(b,c,a),\ $ e.g. $\rm\ \sigma(a-b) = b-c.\ $ Define the norm by $\rm\ N(g) = g\ \sigma g\ \sigma^2 g.\,$ Note that $\rm\,N(\sigma\, g) = N(g)\,$ since $\rm\,\sigma^3 = 1.\,$ Put $\rm\,f = abc = N(a),\,$ so $\rm\,\sigma\, f\, =\, f.\,$ Put $\rm\,g\, =\, a-b.\,$ Subtracting each pair of the given three equations yields the following equations
$$\begin{eqnarray*} \rm f\ &=&\rm\ a\ \dfrac{b-c}{a-b} &=&\rm\ \ \ a\ \ \ \dfrac{\sigma\, g}{g} \\
\rm f\ &=&\rm\ b\ \dfrac{c-a}{b-c} &=&\rm\ \,\sigma\ a\ \ \dfrac{\sigma^2 g}{\sigma\, g} \\
\rm f\ &=&\rm\ c\ \dfrac{a-b}{c-a} &=&\rm\ \sigma^2 a\ \dfrac{\sigma^3\,g}{\sigma^2g}
\end{eqnarray*}$$
Multiplying the above, using $\rm\,\sigma^3g = g,\,$ yields $\rm\,f^{3} = N(a) = f,\,$ so $\rm\,f \ne 0\,$ $\,\Rightarrow\,$ $\rm\,f^{2} = 1.\,$ That is essentially the method employed in said answers. I show that this method amounts to simply taking the norm of the first equation. Notice that the second and third equations result from successively applying $\rm\,\sigma\,$ to the first equation, i.e. the given three equations are equivalent to stating that the first equation is preserved by $\rm\,\sigma\,$ and $\rm\,\sigma^2\,$. So multiplying them all amounts to taking the norm of the first, i.e.
$$\rm  f^{3} = N(f) = N\bigg(a\,\dfrac{\sigma g}{g}\bigg) = N(a) = f\ \quad by\ \quad N(\sigma g) = N(g) $$
So, from this viewpoint, the proof that $\rm\,f^{3} = f\,$ is a one-line inference achieved by by taking a norm (that $\rm\,f/a = \sigma\,g/g\,$ has norm $1$ is essentially multiplicative telescopy). This simplicity above results from recognizing and exploiting the innate symmetry in the problem.
Less trivial exploitation of similar symmetries arise in the Galois theory of difference equations and radical extensions (Kummer theory).  For one simple example see this answer.
A: let $a+\frac{1}{b} = b+\frac{1}{c} = c+\frac{1}{a} = k$
$ab + 1 = bk, bc + 1 = ck, ca + 1 = ak \Rightarrow ab=bk-1, bc=ck-1, ca=ak-1$
$abc + c = bck = (ck-1)k = ck^2-k \Rightarrow abc + k=c(k^2-1)$
$abc + a = ack = (ak-1)k = ak^2-k \Rightarrow abc+k =a(k^2-1)$
$abc+b=abk=(bk-1)k=bk^2-k \Rightarrow abc+k=b(k^2-1)$
$c(k^2-1)=a(k^2-1)=b(k^2-1)=abc + k$
$a\neq b \neq c \neq 0 \Rightarrow k^2-1=0 \Rightarrow k^2=1 \Rightarrow k= \pm1$
$abc+k=c(k^2-1)=0 \Rightarrow abc=-k=\pm1$
A: From the fact that Expressions $1$ and $2$ are equal, we obtain
$$a-b=\frac{1}{c}-\frac{1}{b}=\frac{b-c}{bc}.$$
From the fact that Expressions $2$ and $3$ are equal, we obtain
$$b-c=\frac{1}{a}-\frac{1}{c}=\frac{c-a}{ca}.$$
From the fact that Expressions $3$ and $1$ are equal, we obtain
$$c-a=\frac{1}{b}-\frac{1}{a}=\frac{a-b}{ab}.$$
Multiply the left-hand sides, the right-hand sides.  We get
$$(a-b)(b-c)(c-a)=\frac{(b-c)(c-a)(a-b)}{(abc)^2}.$$
Since $a$, $b$, and $c$ are distinct, $(a-b)(b-c)(c-a)\ne 0$.  We conclude that 
$(abc)^2=1$.  This yields the two possibilities $abc=1$ and $abc=-1$.
In a logical sense we are finished: We have shown that if $(a,b,c)$ is a solution of the system with $a$, $b$, and $c$ distinct, then $abc=\pm 1$.
But it would be nice to show that there are solutions of the desired type.  So let's find such a solution, with $abc=1$.
Look for a solution with $a=1$.  Then we need $c=1/b$.  In order to satisfy our equations, we need $1+1/b=2b$. Beside the useless solution $b=1$, this has the solution $b=-1/2$.  We conclude that 
$$a=1,\quad b=-\frac{1}{2},\quad c=-2$$
is a solution of the desired type, with $abc=1$.  By changing all the signs, we find that 
$$a=-1,\quad b=\frac{1}{2},\quad c=2$$ 
is a solution with $abc=-1$.
Added: It is easy to see that if two 0f $a$, $b$, $c$ are equal, then they are all equal, giving the parametric family $(t,t,t)$, where $t\ne 0$.  Now assume that $abc=\pm 1$.  We find all solutions with $abc=1$. The solutions with $abc=-1$ are then obtained by changing all the signs.  
Let $a=t$.  Our equations will be satisfied if $t+1/b=b+1/c$, where $c=1/bt$. We therefore obtain the equation $t+1/b=b+bt$, which simplifies to $(1+t)b^2 -tb-1=0$. Now we can solve this quadratic equation for $b$, and get $c$ from $tbc=1$. There is undoubtedly a more symmetric way to obtain the complete parametric description of the solutions!
A: Give the common value of $a+1/b$ etc. a name, say $h$. We can now rewrite $a+1/b=h$ to $a=h-1/b$ and similarly $b=h-1/c$ and $c=h-1/a$. Telescoping these expressions into each other gives
$$a=\frac{h^2-ah-1}{h^3-ah^2-2h+a}$$
which rearranges to
$$(1-h^2)a^2 + (h^3-h)a + 1-h^2 = 0$$
Now because everything is symmetric, $b$ and $c$ must satisfy the same equation, but $a$, $b$ and $c$ were distinct numbers, so they can only be roots in a quadratic polynomial if it is identically zero. So $h^2=1$, which makes all of the coefficients vanish. Thus $h=\pm 1$.
Assuming that $h=1$ we now get $b=\frac{1}{1-a}$ and $c=1-\frac{1}{a}=\frac{a-1}{a}$, hence $abc=-1$. An example is $(a,b,c)=(2,-1,1/2)$.
Negating everything in the $h=1$ case gives $h=-1$ and $abc=1$.
A: Let $A = abc$. We can assume $A \ne 0$.
Now the equations can be written as
$a + ac/A = b + ab/A = c + bc/A$
Multiply by $A$ throughout.
$a(A+c) = b(A+a) = c(A+b)$
Subtract first two, last two, first and last to get three equations and multiply those to get an equation in $A$.
A: 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
$(abc)^3 =
 \dfrac{a(b-c)}{a-b} \cdot \dfrac{c(b-a)}{a-c} \cdot \dfrac{b(c-a)}{b-c}$
$(abc)^3 = abc$
$(abc)^2 = 1$
$abc = \pm 1$
$|abc| = 1$
