Cube of an integer $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=k$ and $x, y, z, k$ are integers.
Prove that $xyz$ is cube of some integer number.

I was wondering about giving a parametrization for the rational points on the elliptic curve in $\mathbb{P}^2(\mathbb{R})$
$$ x^2 y + y^2 z + z^2 x = k xyz$$
but I am not too confident in the topic, so I am asking you to shed some light.
Thank you for your help.
 A: Let $p$ be a prime if $p$ divides $x$, $y$ and $z$ then it contributes a cube to $xyz$ and further they can all be divided by $p$ to obtain the same equation again. So assume without loss that $p$ does not divide $z$.
Let $p$ divide $x$ then from the equation in the form 
$$x^2z+y^2x+z^2y=kxyz$$ we see that $p \mid z^2y$ and since by assumption it does not divide $z$ we have $p \mid y$. Let $x=p^na$ and $y=p^mb$ be the largest exponents of $p$ dividing $x$ and $y$.
Then we have 
$$p^{2n}a^2z+p^{2m+n}b^2a+p^mz^2b=kp^{n+m}abz$$ 
Now first if all exponents $2n, 2m+n, m$ are different then their minimum is greater than or equal to $m+n$ (one must admit the possibility that $p$ divides also $k$).
Now $m<2m+n$ so $2m+n$ is not a minimum. The two possible minima are $2n$ and $m$.
If $m$ is the minimum then $n+m\leq m$ and this is impossible.
If $2n$ is the minimum then $n+m\leq 2n$ so $m\leq n$, but if $2n$ is a minimum then $2n < m$ also impossible. This means that there must be two or more equal minimum values.
$2m+n$ is not a minimum are remarked before and thus we are left with $2n=m$.
This implies that $p^{3n}$ is the exact power of $p$ in $xyz$. Therefore $xyz$ is a cube. 
