1981 AHSME Problems/Problem 8

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Problem 8

For all positive numbers $x$, $y$, $z$, the product \[(x+y+z)^{-1}(x^{-1}+y^{-1}+z^{-1})(xy+yz+xz)^{-1}[(xy)^{-1}+(yz)^{-1}+(xz)^{-1}]\] equals

$\textbf{(A)}\ x^{-2}y^{-2}z^{-2}\qquad\textbf{(B)}\ x^{-2}+y^{-2}+z^{-2}\qquad\textbf{(C)}\ (x+y+z)^{-1}\qquad \textbf{(D)}\ \dfrac{1}{xyz}\qquad \\ \textbf{(E)}\ \dfrac{1}{xy+yz+xz}$


Solution

Start simplifying: \[(\frac{1}{x+y+z})(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(\frac{1}{xy+yz+xz})(\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz})\]

\[(\frac{1}{x+y+z})(\frac{xy+yz+xz}{xyz})(\frac{1}{xy+yz+xz})(\frac{x+y+z}{xyz})\]

The $xy+yz+xz$ and $x+y+z$ cancel out:

\[\frac{1}{(xyz)^2}\]

The answer is $\textbf{(A)}\ x^{-2}y^{-2}z^{-2}.$

-edited by coolmath34