1990 AIME Problems/Problem 2
Contents
Problem
Find the value of .
Solution 1
Suppose that is in the form of . FOILing yields that . This implies that and equal one of . The possible sets are and ; the latter can be discarded since the square root must be positive. This means that . Repeating this for , the only feasible possibility is .
Rewriting, we get . Using the difference of cubes, we get that .
Solution 2
The power is quite irritating to work with so we look for a way to eliminate that. Notice that squaring the expression will accomplish that. Let be the sum of the given expression. After doing the arithmetic (note that the first two terms will have some cancellation and that the last term will simplify quickly using difference of squares), we arrive at which gives .
Solution 3
Factor as a difference of cubes. We can simplify the left factor as follows. Since , we know that , so our final answer is .
Solution 4
Let , . Similarly to solution 2, we let The expression can be simplified as follow Thus .
~ Nafer
Solution 5
(Similar to Solution 3, but with substitution)
Let and We want to find
We have Then,
Our answer is
Video Solution
https://www.youtube.com/watch?v=r96p8j0F8Fg
See also
1990 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.