Difference between revisions of "2008 AIME II Problems/Problem 7"
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<cmath>\begin{align*}8\{(r + s)^3 + (s + t)^3 + (t + r)^3\} &= - 8(r^3 + s^3 + t^3)\\ | <cmath>\begin{align*}8\{(r + s)^3 + (s + t)^3 + (t + r)^3\} &= - 8(r^3 + s^3 + t^3)\\ | ||
&= 1001(r + s + t) + 2008\cdot 3 = 3\cdot 2008\end{align*}</cmath>yielding the answer <math>753</math>. | &= 1001(r + s + t) + 2008\cdot 3 = 3\cdot 2008\end{align*}</cmath>yielding the answer <math>753</math>. | ||
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+ | Also, Newton's Sums yields an answer through the application. | ||
+ | http://www.artofproblemsolving.com/Wiki/index.php/Newton's_Sums | ||
== See also == | == See also == |
Revision as of 17:35, 11 August 2014
Problem
Let ,
, and
be the three roots of the equation
Find
.
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
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.