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For how many ordered triples of unequal positive integers $(x,y,z)$ does the expression $$ \frac{x}{(x-y)(x-z)} + \frac{y}{(y-x)(y-z)} + \frac{z}{(z-x)(z-y)} $$ take on positive values?

I started with $x=3, y=4$ and $z=5$ and got:

$\frac{3}{2}+\frac{4}{-1}+\frac{5}{2}=0$

Then I worked with $x=1, y=2$ and $z=3$

$\frac{1}{2}+\frac{2}{-1}+\frac{3}{2}=0$

Then I worked with: $x=4, y=5$ and $z=6$

$\frac{4}{2}+\frac{5}{-1}+\frac{6}{2}=0$

So I did a random triple: $x=8, y=9$ and $z=10$

$\frac{8}{2}+\frac{9}{-1}+\frac{10}{2}=0$

So the answer to this question would be none, but I could be wrong. Any ideas?

mvw
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sharonnicole116
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1 Answers1

4

$\dfrac{x}{(x-y)(x-z)} + \dfrac{y}{(y-x)(y-z)} + \dfrac{z}{(z-x)(z-y)}$

$\dfrac{x}{(x-y)(x-z)} + \dfrac{-y}{(x-y)(y-z)} + \dfrac{z}{(x-z)(y-z)}$

$\dfrac{x(y-z) -y(x-z) + z(x-y)}{(x-y)(x-z)(y-z)}= \dfrac{0}{(x-y)(x-z)(y-z)} = 0$

IrbidMath
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