Difference between revisions of "2008 AIME II Problems/Problem 7"
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Find <math>(r + s)^3 + (s + t)^3 + (t + r)^3</math>. | Find <math>(r + s)^3 + (s + t)^3 + (t + r)^3</math>. | ||
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+ | ==Video Solution by Punxsutawney Phil== | ||
+ | https://www.youtube.com/watch?v=6mYZYh9gJBs | ||
==Solution 1 == | ==Solution 1 == |
Revision as of 09:47, 5 July 2022
Contents
[hide]Problem
Let ,
, and
be the three roots of the equation
Find
.
Video Solution by Punxsutawney Phil
https://www.youtube.com/watch?v=6mYZYh9gJBs
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.