Difference between revisions of "2008 AIME II Problems/Problem 7"
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<math>(r+s)^3+(s+t)^3+(t+r)^3=3\cdot\frac{2008}{8}-\frac{1001}{8}(2r+2s+2t)=251(3)+0=\boxed{753}</math> | <math>(r+s)^3+(s+t)^3+(t+r)^3=3\cdot\frac{2008}{8}-\frac{1001}{8}(2r+2s+2t)=251(3)+0=\boxed{753}</math> | ||
+ | ==Solution 8== | ||
+ | |||
+ | We want to find what is <math>-(r^3+s^3+t^3)</math> which reminds us of Newton sum. So we can see that <math>8S_3+0\cdot S_2+1001\cdot S_1+3\cdot 2008=0</math> Notice that <math>S_1=0</math> so it is just <math>S_3=-\frac{2008\cdot 3}{8}=-753</math>, the desired answer is <math>\boxed{753}</math> | ||
+ | |||
+ | ~bluesoul | ||
== See also == | == See also == | ||
{{AIME box|year=2008|n=II|num-b=6|num-a=8}} | {{AIME box|year=2008|n=II|num-b=6|num-a=8}} |
Revision as of 13:39, 24 June 2022
Contents
[hide]Problem
Let , , and be the three roots of the equation Find .
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
Solution 7
Let's construct a polynomial with the roots and .
sum of the roots:
pairwise product of the roots:
product of the roots:
thus, the polynomial we get is
as and are roots of this polynomial, we know that (using power reduction)
adding all of the equations up, we see that
Solution 8
We want to find what is which reminds us of Newton sum. So we can see that Notice that so it is just , the desired answer is
~bluesoul
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.