<s>[INST] Prove that if $x$, $y$, and $z$ are real numbers such that $x^2(y-z)+y^2(z-x)+z^2(x-y)=0,$ then at least two of them are equal 
Prove that if $x$, $y$, and $z$ are real numbers such that $x^2(y-z)+y^2(z-x)+z^2(x-y)=0,$ then at least two of them are equal

This question was asked in a past international math exam. My idea to solve it was to factorize it and then end up showing that the polynomial that was formed would directly show that at least two of the numbers $x$, $y$, $z$ are equal. However, it was to no avail.
I researched for theorems, that may help me in my quest to solving this problem, but none of them seemed applicable to this problem. Can someone please show me, a method of either finishing my thought, or if you believe that I was on the completely wrong trail of thought, can you please show me another method of solving it?
 [/INST] Factoring is hard and I'm not good at it.
But if:
$x^2(y-z)+y^2(z-x) + z^2(x-y) = 0$
$x^2(y-z)+y^2(z-x)= z^2(y-x)$ 
If $y =x$ then $y=x$.  But if $y\ne x$ then $y-x \ne 0$.
$\frac {x^2(y-z) + y^2(z-x)}{y-x} = z^2$
So $\frac {xy*y - x^2z + y^2z - xy*y}{y-x} =$
$ -xy + \frac{y^2z -x^2z}{y-x} = $
$-xy + z\frac{(y-x)(y+x)}{y-x} =$
$-xy + z(y+x)=$
$-xy + zy + xz = z^2$
And so
$xz -xy = z^2 - zy$
$x(z-y) = z(z-y)$.  
If $z = y$ then $z=y$ but if $z \ne y$ the  $z - y\ne 0$ and
$x = z$.
So either $x=y$ or if not then either $z=y$ or if not $x = z$.
....
Which actually helps me figure out how to factor.  So we could have done the following instead:
We can factor $x-y$ (or equivalently $y-x$) from $x^2(y-z) + y^2(z-x)$ via 
$x^2(y-z) + y^2(z-x) =$
$xy*x - xy*y - (z*x^2 - z*y^2) = $
$(x-y)(xy - z(x+y))$
So $x^2(y-z) + y^2(z-x) + z^2(x-y) = $
$(x-y) (xy -z(x+y) + z^2)$
And we can factor $y-z$ from $xy - z(x+y) + z^2$ via
$xy - z(x+y) + z^2 = $
$xy - zx + z^2 - zy =$
$x(y-z) + z(z-y) =$
$(y-z)(x-z)$
And thus:
$x^2(y-z) + y^2(z-x) + z^2(x-y)= (x-y)(y-z)(x-z) = 0$
.... which means one of $x-y$ or $y-z$ or $x - z=0$.
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