Difference between revisions of "2008 AIME II Problems/Problem 7"
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We are trying to find <math> -(r^3 + s^3 + t^3)</math>. | We are trying to find <math> -(r^3 + s^3 + t^3)</math>. | ||
Substituting: | Substituting: | ||
− | <cmath> -(r^3 + s^3 + t^3) = -3srt = \frac{-2008*3}{8} = \boxed{753}</cmath> | + | <cmath> -(r^3 + s^3 + t^3) = -3srt = \frac{-2008*3}{8} = \boxed{753}.</cmath> |
=== Solution 4 === | === Solution 4 === |
Revision as of 09:38, 31 May 2018
Problem
Let , , and be the three roots of the equation Find .
Contents
[hide]Solution
Solution 1
By Vieta's formulas, we have , and so the desired answer is . Additionally, using the factorization we have that . By Vieta's again,
Solution 2
Vieta's formulas gives . Since is a root of the polynomial, , and the same can be done with . Therefore, we have yielding the answer .
Also, Newton's Sums yields an answer through the application. http://www.artofproblemsolving.com/Wiki/index.php/Newton's_Sums
Solution 3
Expanding, you get: This looks similar to Substituting: Since , Substituting, we get or, We are trying to find . Substituting:
Solution 4
Write and let . Then Solving for and negating the result yields the answer
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.