If $a+\frac1b=b+\frac1c=c+\frac1a$ for distinct $a$, $b$, $c$, how to find the value of $abc$?

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!

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 $abc=1$.

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 $A$.