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

(Solution)
Line 17: Line 17:
 
<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>.
 
 
  
 
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 $r$, $s$, 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 $r+s+t = 0$, 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 2

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 $753$.

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: \[r^3 + 3r^2s + 3s^2r +s^3 +\] \[s^3 + 3s^2t + 3t^2s +t^3 +\] \[r^3 + 3r^2t + 3t^2r +t^3\] \[= 2r^3 + 2s^3 + 2t^3 + 3r^2s + 3s^2r + 3s^2t + 3t^2s + 3r^2t + 3t^2r\] 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: \[-(r^3 + s^3 + t^3) = -3srt = \frac{-2008*3}{8} = \boxed{753}\].

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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