Difference between revisions of "2008 AIME II Problems/Problem 7"
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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 | 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)\ | <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>\boxed{753}</math>. |
Also, Newton's Sums yields an answer through the application. | Also, Newton's Sums yields an answer through the application. | ||
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Substituting: | Substituting: | ||
<cmath> -(r^3 + s^3 + t^3) = -3srt = \frac{-2008*3}{8} = \boxed{753}</cmath>. | <cmath> -(r^3 + s^3 + t^3) = -3srt = \frac{-2008*3}{8} = \boxed{753}</cmath>. | ||
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+ | === 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> | ||
+ | |||
== 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 09:38, 31 May 2018
Problem
Let , , and be the three roots of the equation Find .
Contents
[hide]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 |
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