1981 AHSME Problems/Problem 8
Problem 8
For all positive numbers 
, 
, 
, the product
![\[(x+y+z)^{-1}(x^{-1}+y^{-1}+z^{-1})(xy+yz+xz)^{-1}[(xy)^{-1}+(yz)^{-1}+(xz)^{-1}]\]](http://latex.artofproblemsolving.com/b/c/7/bc79c652eb33e7237474c872a5fa9dcb78c0dd79.png)
equals
Solution
Converting all negative exponents to fractions yields:
![\[\left(\frac{1}{x+y+z}\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{xy+yz+xz}\right)\left(\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}\right)\]](http://latex.artofproblemsolving.com/c/e/8/ce84b57ed50231b423037cc039af647038f299b5.png)
Adding fractions within each factor yields:
![\[\left(\frac{1}{x+y+z}\right)\left(\frac{xy+yz+xz}{xyz}\right)\left(\frac{1}{xy+yz+xz}\right)\left(\frac{x+y+z}{xyz}\right)\]](http://latex.artofproblemsolving.com/6/6/f/66fb917ac1c0af04600ad82b6cd5d3dd803247b9.png)
Multiplying and cancelling common factors yields:
![\[\frac{1}{x^2 y^2 z^2} = \boxed{\textbf{(A)}\ x^{-2}y^{-2}z^{-2}}\]](http://latex.artofproblemsolving.com/4/3/0/4305e29e2efcb787a532283499b2230dea65a968.png)
-edited by coolmath34
See also
| 1981 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
