# Proving the independence of an equation, from certain variables

The different and unequal to zero real numbers x, y, z satisfy the equation

$$x^3+y^3+m(x+y)=y^3+z^3+m(y+z)=z^3+x^3+m(z+x).$$

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 $$x, y, z, m$$.

I don't know how to solve the question above. I've been trying it for a while, but I can't. Can you guys please help me?

Kevin

• What's the source of the problem? – Dr. Mathva Apr 14 at 18:33
• a leaflet given for the preperation for the international exam of JBMO – kenith Apr 14 at 18:34

$$x^3 + y^3 + m(x+y) = y^3 + z^3 + m(y+z) = z^3 + x^3 + m(z+x)$$

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.