Difference between revisions of "2008 AIME II Problems/Problem 7"
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<cmath>r^3 + 3r^2t + 3t^2r +t^3</cmath> | <cmath>r^3 + 3r^2t + 3t^2r +t^3</cmath> | ||
<cmath>= 2r^3 + 2s^3 + 2t^3 + 3r^2s + 3s^2r + 3s^2t + 3t^2s + 3r^2t + 3t^2r </cmath> | <cmath>= 2r^3 + 2s^3 + 2t^3 + 3r^2s + 3s^2r + 3s^2t + 3t^2s + 3r^2t + 3t^2r </cmath> | ||
− | This looks similar to <math>(r+s+t)^3 = r^3 + s^3 + t^3 + 3r^2s + 3s^2r + 3s^2t + 3t^2s + 3r^2t + 3t^2r + | + | This looks similar to <math>(r+s+t)^3 = r^3 + s^3 + t^3 + 3r^2s + 3s^2r + 3s^2t + 3t^2s + 3r^2t + 3t^2r + 6rst</math> |
Substituting: | Substituting: | ||
<cmath>(r+s+t)^3 - 6rst + r^3+s^3+t^3 = (r + s)^3 + (s + t)^3 + (t + r)^3</cmath> | <cmath>(r+s+t)^3 - 6rst + r^3+s^3+t^3 = (r + s)^3 + (s + t)^3 + (t + r)^3</cmath> |
Revision as of 18:01, 28 May 2020
Problem
Let , , and be the three roots of the equation Find .
Contents
[hide]Solution
Solution 1
By Vieta's formulas, we have so Substituting this into our problem statement, our desired quantity is Also by Vieta's formulas we have so negating both sides and multiplying through by 3 gives our answer of
Solution 2
By Vieta's formulas, we have , and so the desired answer is . Additionally, using the factorization we have that . By Vieta's again,
Solution 3
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 4
Expanding, you get: This looks similar to Substituting: Since , Substituting, we get or, We are trying to find . Substituting:
Solution 5
Write and let . Then Solving for and negating the result yields the answer
Solution 6
Here by Vieta's formulas: --(1)
--(2)
By the factorisation formula: Let , , , (By (1))
So
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.