1981 AHSME Problems/Problem 8

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

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1981 AHSME Problems/Problem 8

Problem 8

Solution