Proving the independence of an equation, from certain variables 
Assume that $x, y, z$ are three distinct nonzero real numbers satisfying the equation
$$x^3+y^3+m(x+y)=y^3+z^3+m(y+z)=z^3+x^3+m(z+x)$$
for some real $m$. Prove that
$$K=\left(\frac{x-y}{z}+\frac{y-z}{x}+\frac{z-x}{y}\right)\left(\frac{z}{x-y}+\frac{x}{y-z}+\frac{y}{z-x}\right)$$
is not dependent upon $x, y, z, m$.

This problem comes from a leaflet given for the preparation for the international exam of JBMO. I don't know how to solve it. I've been trying it for a while, but I can't.
I know that the equation
$$x^3+y^3+m(x+y)=y^3+z^3+m(y+z)=z^3+x^3+m(z+x)$$
can be rewritten as
$$m(x+y)-z^3=m(y+z)-x^3=m(z+x)-y^3 ,$$
which is shorter but not necessarily more useful.
 A: $$
x^3 + y^3 + m(x+y) = y^3 + z^3 + m(y+z) = z^3 + x^3 + m(z+x)
$$
holds for some $m \in \mathbb R$, if and only if
$$
\begin{cases}
x^3 - y^3 = -m(x-y) \\
y^3 - z^3 = -m(y-z)
\end{cases}
$$
holds for some $m \in \mathbb R$. If we presume that $x,y,z$ are pairwise distinct, this is equivalent to saying that, considering the curve $Y = X^3$ on the plane, the three points $(x, x^3), (y, y^3)$ and $(z, z^3)$ are co-linear.
Now we look a bit geometrically. If the line $Y = aX + b$ intersects $Y = X^3$ at three points $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$, then $x_1, x_2, x_3$ are the roots of the equation $X^3 = aX + b$, so $-x_1-x_2-x_3$ are the $x^2$ coefficient, which is zero. So if we move back to our problem, this is equivalent to saying that
$$
x+y+z=0.
$$
The second half of the problem is to show that
$$
K = \left(\frac{x-y}z + \frac{y-z}x + \frac{z-x}y\right)\left(\frac z{x-y} + \frac x{y-z} + \frac y{z-x}\right)
$$
is constant whenever $x+y+z=0$. You need to know that the constant is, if such constant exists. Taking $(x,y,z) = (1,-3,2)$ for instance, you see that $K$ is $c = 9$. So we need to show that $K$ is $c=9$ all the time.
Observe that as $K$ is homogeneous of degree 0, multiplying $x,y,z$ simultaneously by a constant $c$ does not change $K$. Therefore we can scale $x$ to $1$, assume $y = t$ and $z = -t-1$. This gives us an expression
$$
K = \frac{p(t)}{q(t)},
$$
where $p(t)$ and $q(t)$ has degree at most6 and 5 in $t$, respectively.
To show that $K$ is constantly 9, you need to show that $F(t) = p(t) - 9q(t)$ is constantly zero. You can expand it, which will be brutal; an easier way is to observe that $F$ is of degree at most 6, so if you can find seven different $t$ such that $F(t) = 0$, then we are done. The pair $(1,-3,2)$ we started with, we see that $F(-3) = 0$, with a bit more insight we also see that $F(-2/3) = F(1/2) = 0$ (why?), and we can also tell that $F(2) = F(-1/3) = F(-3/2) = 0$. Now you get six zeros, and try a new set of test value to get the seventh.
