Difference between revisions of "2008 AIME II Problems/Problem 7"
(→Solution) |
(→Solution 4) |
||
Line 39: | Line 39: | ||
=== Solution 4 === | === Solution 4 === | ||
− | Write <math>(r+s)^3+(s+t)^3+(t+r)^3=-(r^3+s^3+t^3)</math> and let <math>f(x)=8x^3+1001x+2008</math>. Then <cmath>f(r)+f(s)+f(t)=8(r^3+s^3+t^3)+1001(r+s+t)+6024=8(r^3+s^3+t^3)+6024=0.</cmath> Solving for <math>r^3+s^3+t^3</math> and negating the result yields the answer <math>\boxed{753}</math> | + | Write <math>(r+s)^3+(s+t)^3+(t+r)^3=-(r^3+s^3+t^3)</math> and let <math>f(x)=8x^3+1001x+2008</math>. Then <cmath>f(r)+f(s)+f(t)=8(r^3+s^3+t^3)+1001(r+s+t)+6024=8(r^3+s^3+t^3)+6024=0.</cmath> Solving for <math>r^3+s^3+t^3</math> and negating the result yields the answer <math>\boxed{753}.</math> |
== See also == | == See also == |
Revision as of 10:38, 31 May 2018
Problem
Let ,
, and
be the three roots of the equation
Find
.
Contents
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.