Difference between revisions of "2008 AIME II Problems/Problem 7"

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===Solution 1 ===
 
===Solution 1 ===
By [[Vieta's formulas]], we have <math>r + s + t = 0</math> so <math>t = -r - s.</math> Substituting this into our problem statement, our desired quantity is <cmath>(r + s)^3 - r^3 - s^3 = 3r^s + 3rs^2 = 3rs(r + s).</cmath> Also by [[Vieta's formulas]] we have <cmath>rst = -rs(r + s) = -\dfrac{2008}{8} = -251</cmath> so negating both sides and multiplying through by 3 gives our answer of <math>\boxed{753}.</math>
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By [[Vieta's formulas]], we have <math>r + s + t = 0</math> so <math>t = -r - s.</math> Substituting this into our problem statement, our desired quantity is <cmath>(r + s)^3 - r^3 - s^3 = 3r^2s + 3rs^2 = 3rs(r + s).</cmath> Also by [[Vieta's formulas]] we have <cmath>rst = -rs(r + s) = -\dfrac{2008}{8} = -251</cmath> so negating both sides and multiplying through by 3 gives our answer of <math>\boxed{753}.</math>
  
 
=== Solution 2 ===
 
=== Solution 2 ===

Revision as of 13:38, 27 March 2020

Problem

Let $r$, $s$, and $t$ be the three roots of the equation \[8x^3 + 1001x + 2008 = 0.\] Find $(r + s)^3 + (s + t)^3 + (t + r)^3$.

Solution

Solution 1

By Vieta's formulas, we have $r + s + t = 0$ so $t = -r - s.$ Substituting this into our problem statement, our desired quantity is \[(r + s)^3 - r^3 - s^3 = 3r^2s + 3rs^2 = 3rs(r + s).\] Also by Vieta's formulas we have \[rst = -rs(r + s) = -\dfrac{2008}{8} = -251\] so negating both sides and multiplying through by 3 gives our answer of $\boxed{753}.$

Solution 2

By Vieta's formulas, we have $r+s+t = 0$, and so the desired answer is $(r+s)^3 + (s+t)^3 + (t+r)^3 = (0-t)^3 + (0-r)^3 + (0-s)^3 = -(r^3 + s^3 + t^3)$. Additionally, using the factorization \[r^3 + s^3 + t^3 - 3rst = (r+s+t)(r^2 + s^2 + t^2 - rs - st - tr) = 0\] we have that $r^3 + s^3 + t^3 = 3rst$. By Vieta's again, $rst = \frac{-2008}8 = -251 \Longrightarrow -(r^3 + s^3 + t^3) = -3rst = \boxed{753}.$

Solution 3

Vieta's formulas gives $r + s + t = 0$. Since $r$ is a root of the polynomial, $8r^3 + 1001r + 2008 = 0\Longleftrightarrow - 8r^3 = 1001r + 2008$, and the same can be done with $s,\ t$. Therefore, we have \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*}yielding the answer $\boxed{753}$.

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: \[r^3 + 3r^2s + 3s^2r +s^3 +\] \[s^3 + 3s^2t + 3t^2s +t^3 +\] \[r^3 + 3r^2t + 3t^2r +t^3\] \[= 2r^3 + 2s^3 + 2t^3 + 3r^2s + 3s^2r + 3s^2t + 3t^2s + 3r^2t + 3t^2r\] This looks similar to $(r+s+t)^3 = r^3 + s^3 + t^3 + 3r^2s + 3s^2r + 3s^2t + 3t^2s + 3r^2t + 3t^2r + rst$ Substituting: \[(r+s+t)^3 - 6rst + r^3+s^3+t^3 = (r + s)^3 + (s + t)^3 + (t + r)^3\] Since $r+s+t = 0$, \[(r+s)^3 + (s+t)^3 + (t+r)^3 = (0-t)^3 + (0-r)^3 + (0-s)^3 = -(r^3 + s^3 + t^3)\] Substituting, we get \[(r+s+t)^3 - 6rst + r^3+s^3+t^3 =  -(r^3 + s^3 + t^3)\] or, \[0^3 - 6rst + r^3+s^3+t^3 =  -(r^3 + s^3 + t^3) \implies 2(r^3 + s^3 + t^3) = 6rst\] We are trying to find $-(r^3 + s^3 + t^3)$. Substituting: \[-(r^3 + s^3 + t^3) = -3srt = \frac{-2008*3}{8} = \boxed{753}.\]

Solution 5

Write $(r+s)^3+(s+t)^3+(t+r)^3=-(r^3+s^3+t^3)$ and let $f(x)=8x^3+1001x+2008$. Then \[f(r)+f(s)+f(t)=8(r^3+s^3+t^3)+1001(r+s+t)+6024=8(r^3+s^3+t^3)+6024=0.\] Solving for $r^3+s^3+t^3$ and negating the result yields the answer $\boxed{753}.$

Solution 6

Here by Vieta's formulas: $r+s+t = 0$ --(1)

$rst = \frac{-2008}{8} = -251$ --(2)

By the factorisation formula: Let $a = r+s$, $b = s+t$, $c = t+r$, $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = 0$ (By (1))

So \[a^3+b^3+c^3 = 3abc = 3(r+s)(s+t)(t+r) = 3(-t)(-r)(-s) = 3[-(-251)] = \boxed{753}.\]

See also

2008 AIME II (ProblemsAnswer KeyResources)
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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