For what real numbers $x, y, z \not \in \{0, 1\},$ does the following hold true?
$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$$
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The set $\{ x,y,z \, | \, 1/x + 1/y + 1/z = 1 \} \subset \mathbb{R}^3$ describes a two-dimensional surface in $\mathbb{R}^3$. You could for example parametrize it by $x$ and $y$, such that $z$ becomes a function of $x$ and $y$: $$z = f(x,y).$$ In order to find $f(x,y)$, you can just manipulate the above equation - I'll leave that up to you, since it's homework.
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$\begingroup$ I'm looking for something a 9th grader could solve. It can be done with parametrization, but is there a more simplistic way? $\endgroup$ Sep 18, 2011 at 10:31
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1$\begingroup$ @Paul: There's already on the table here an implicit equation and an explicit parametrization; what other ways to represent a surface are there? The only other thing I can think of is that they will be the real roots of polynomials of the form $u^3-au^2+bu-b$. $\endgroup$– anonSep 18, 2011 at 11:24