$x + \frac1y = y + \frac1z = z + \frac1x$, then value of $xyz$ is? If $x,y,z$ are distinct positive numbers, such that $$x + \frac1y = y + \frac1z = z + \frac1x $$ then value of $xyz$ is?
$$A)\ 4\quad  B)\ 3\quad C)\ 2\quad  D)\ 1$$
My attempt: 
1.I equaled the equation to '$k$'. Using the AM-GM inequality, I found that $k>2$ (the equality does not hold because all are distinct). However, this, I couldn't put to much use.
2. For the next attempt, I substituted the values of $x$ and $y$ in terms of $z$ and $k$ in $xyz$. What I am getting is $xyz=zk^2 - k - z$. That's the farthest I could do..Please, help.
 A: Other answers have shown that the question is flawed. For fun, let's remove the condition that $x,y,z$ are positive, but keep the requirement that $x,y,z$ are distinct, and try to find all solutions $(x,y,z)$.
Trivially, $x,y,z \neq 0$. If we set $x+\dfrac{1}{y} = y+\dfrac{1}{z} = z + \dfrac{1}{x} = k$, for som real number $k$, then 
\begin{align}
x &= k - \dfrac{1}{y}
\\
y &= k - \dfrac{1}{z}
\\
z &= k - \dfrac{1}{x}
\end{align}
If we substitute equation (3) into (2) and then substitute that into (1), we get $$x = k-\dfrac{1}{k-\dfrac{1}{k-\tfrac{1}{x}}}.$$
By unraveling this fraction, we eventually get the equation $$(k^2-1)(x^2-kx+1) = 0.$$
If $k \neq \pm 1$, then the solutions $x$ will satisfy $$x^2-kx+1 = 0 \iff x = k - \dfrac{1}{x} = z,$$ and we won't have distinctness. 
If $k = 1$, then $z = 1-\dfrac{1}{x} = \dfrac{x-1}{x}$ and $y = 1-\dfrac{1}{z} = -\dfrac{1}{x-1}$. 
So the solutions in this case are $(x,y,z) = \left(x,-\dfrac{1}{x-1},\dfrac{x-1}{x}\right)$, all of which satisfy $xyz = -1$.
If $k = -1$, then $z = -1-\dfrac{1}{x} = -\dfrac{x+1}{x}$ and $y = -1-\dfrac{1}{z} = -\dfrac{1}{x+1}$.
So the solutions in this case are $(x,y,z) = \left(x,-\dfrac{1}{x+1},-\dfrac{x+1}{x}\right)$, all of which satisfy $xyz = 1$.
Therefore, the only solutions where $x,y,z$ are distinct are all of the form $(x,y,z) = \left(x,-\dfrac{1}{x-1},\dfrac{x-1}{x}\right)$ for $x \neq 0,1$ or $(x,y,z) = \left(x,-\dfrac{1}{x+1},-\dfrac{x+1}{x}\right)$ for $x \neq 0,-1$. (It's easy to check that these satisfy the distinctness condition.) Also, the only possible values of $xyz$ are $\pm 1$.
A: By multiplying the first equation by $yz$, the second by $xz$, and the equation connecting the first and third expression by $xy$, we get
$$xyz+z = y^2z+y$$
$$xyz+x = z^2x+z$$
$$xyz+y = x^2y+x$$
Adding these three, cancelling $x+y+z$ from both sides, and dividing by $xyz$ gives $$\frac{x}{z}+\frac{y}{x}+\frac{z}{y} = 3$$
But by AM-GM, the LHS is $\ge 3$ and so we must have equality in AM-GM, meaning $\frac{x}{z} = \frac{y}{x} = \frac{z}{y}$, and so $x^2 = yz, y^2 = xz, z^2 = xy$. Dividing the first by the second gives $\frac{x^2}{y^2} = \frac{y}{x}$ and so $x^3 = y^3 \implies x=y$. Similarly, $y=z$. This contradicts that $x,y,z$ are distinct, so the question is flawed.
A: From the conditions we have:
$$x-y=\frac{y-z}{yz}$$ $$z-x=\frac{x-y}{xy}$$ and $$y-z=\frac{z-x}{xz}$$
Thus, $xyz=1$, but why it  holds? 
A: Note that $x-y=\frac{y-z}{yz}$, $y-z=\frac{z-x}{zx}$ and $z-x=\frac{x-y}{xy}$. Multiply to give $(x-y)(y-z)(z-x)=\frac{(x-y)(y-z)(z-x)}{(xyz)^2}$. As $x$, $y$ and $z$ distinct this means $(xyz)^2=1$. As $x$, $y$ and $z$ positive we must have $xyz=1$.
A: Since there is a question,
there is an answer.
By inspection,
$x=y=z$
satisfies the equations.
Therefore
$xyz = x^3$.
But the only cube listed is
$1$,
so that must be the answer.
Meta-reasoning strikes again!
