Difference between revisions of "2008 AIME II Problems/Problem 7"
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=== Solution 5 === | === Solution 5 === | ||
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> | ||
+ | |||
+ | === Solution 6 === | ||
+ | Here by [[Vieta's formulas]]: | ||
+ | <math>r+s+t = 0</math> --(1) | ||
+ | |||
+ | <math>rst = \frac{-2008}{8} = -251</math> --(2) | ||
+ | |||
+ | By the factorisation formula: | ||
+ | Let <math>a = r+s</math>, <math>b = s+t</math>, <math>c = t+r</math>, | ||
+ | <math>a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = 0</math> (By (1)) | ||
+ | |||
+ | So <cmath>a^3+b^3+c^3 = 3abc = 3(r+s)(s+t)(t+r) = 3(-t)(-r)(-s) = 3[-(-251)] = \boxed{753}.</cmath> | ||
== See also == | == See also == |
Revision as of 10:45, 5 February 2019
Problem
Let , , and be the three roots of the equation Find .
Contents
Solution
Solution 1
By Vieta's formulas, we have so Substituting this into our problem statement, our desired quantity is Also by Vieta's formulas we have so negating both sides and multiplying through by 3 gives our answer of
Solution 2
By Vieta's formulas, we have , and so the desired answer is . Additionally, using the factorization we have that . By Vieta's again,
Solution 3
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 4
Expanding, you get: This looks similar to Substituting: Since , Substituting, we get or, We are trying to find . Substituting:
Solution 5
Write and let . Then Solving for and negating the result yields the answer
Solution 6
Here by Vieta's formulas: --(1)
--(2)
By the factorisation formula: Let , , , (By (1))
So
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.