Difference between revisions of "2008 AIME II Problems/Problem 7"
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=== Solution 2 === | === Solution 2 === | ||
− | Vieta's formulas gives <math>r + s + t = 0</math>. Since <math>r</math> is a root of the polynomial, <math>8r^3 + 1001r + 2008 = 0\Longleftrightarrow - 8r^3 = 1001r + 2008</math>, and the same can be done with <math>s,\ t</math>. | + | Vieta's formulas gives <math>r + s + t = 0</math>. Since <math>r</math> is a root of the polynomial, <math>8r^3 + 1001r + 2008 = 0\Longleftrightarrow - 8r^3 = 1001r + 2008</math>, and the same can be done with <math>s,\ t</math>. Therefore, we have |
− | < | + | <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 | + | &= 1001(r + s + t) + 2008\cdot 3 = 3\cdot 2008\end{align*}</cmath>yielding the answer <math>753</math>. |
== See also == | == See also == |
Revision as of 18:00, 3 April 2008
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
.
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 |