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. | Also, Newton's Sums yields an answer through the application. | ||
http://www.artofproblemsolving.com/Wiki/index.php/Newton's_Sums | http://www.artofproblemsolving.com/Wiki/index.php/Newton's_Sums | ||
+ | === Solution 3 === | ||
+ | Expanding, you get: | ||
+ | <cmath>r^3 + 3r^2s + 3s^2r +s^3 +</cmath> | ||
+ | <cmath>s^3 + 3s^2t + 3t^2s +t^3 +</cmath> | ||
+ | <cmath>r^3 + 3r^2t + 3t^2r +t^3</cmath> | ||
+ | <cmath>= 2r^3 + 2s^3 + 2t^3 + 3r^2s + 3s^2r + 3s^2t + 3t^2s + 3r^2t + 3t^2r </cmath> | ||
+ | This looks similar to <math>(r+s+t)^3 = r^3 + s^3 + t^3 + 3r^2s + 3s^2r + 3s^2t + 3t^2s + 3r^2t + 3t^2r + rst</math> | ||
+ | Substituting: | ||
+ | <cmath>(r+s+t)^3 - 6rst + r^3+s^3+t^3 = (r + s)^3 + (s + t)^3 + (t + r)^3</cmath> | ||
+ | Since <math>r+s+t = 0</math>, | ||
+ | <cmath>(r+s)^3 + (s+t)^3 + (t+r)^3 = (0-t)^3 + (0-r)^3 + (0-s)^3 = -(r^3 + s^3 + t^3)</cmath> | ||
+ | Substituting, we get <cmath>(r+s+t)^3 - 6rst + r^3+s^3+t^3 = -(r^3 + s^3 + t^3)</cmath> | ||
+ | or, <cmath>0^3 - 6rst + r^3+s^3+t^3 = -(r^3 + s^3 + t^3) \implies 2(r^3 + s^3 + t^3) = 6rst</cmath> | ||
+ | We are trying to find <math> -(r^3 + s^3 + t^3)</math>. | ||
+ | Substituting: | ||
+ | <cmath> -(r^3 + s^3 + t^3) = -3srt = \frac{-2008*3}{8} = \boxed{753}</cmath>. | ||
== 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 22:19, 2 March 2015
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
Solution 3
Expanding, you get: This looks similar to Substituting: Since , Substituting, we get or, We are trying to find . Substituting: .
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 |
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