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

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Find <math>(r + s)^3 + (s + t)^3 + (t + r)^3</math>.
 
Find <math>(r + s)^3 + (s + t)^3 + (t + r)^3</math>.
  
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__TOC__
 
== Solution ==
 
== Solution ==
By [[Vieta’s sums]], we have <math>r+s+t = 0</math>, and so the desired answer is <math>(r+s)^3 + (s+t)^3 + (t+r)^3 = -(r^3 + s^3 + t^3)</math>. Additionally, using the factorization
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=== Solution 1 ===
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By [[Vieta's formulas]], we have <math>r+s+t = 0</math>, and so the desired answer is <math>(r+s)^3 + (s+t)^3 + (t+r)^3 = (0-t)^3 + (0-r)^3 + (0-s)^3 = -(r^3 + s^3 + t^3)</math>. Additionally, using the factorization
 
<cmath>r^3 + s^3 + t^3 - 3rst = (r+s+t)(r^2 + s^2 + t^2 - rs - st - tr) = 0</cmath>
 
<cmath>r^3 + s^3 + t^3 - 3rst = (r+s+t)(r^2 + s^2 + t^2 - rs - st - tr) = 0</cmath>
Thus <math>r^3 + s^3 + t^3 = 3rst</math>. By Vieta’s again, we have <math>rst = \frac{-2008}8 = -251 \Longrightarrow -(r^3 + s^3 + t^3) = -3rst = \boxed{753}.</math>
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we have that <math>r^3 + s^3 + t^3 = 3rst</math>. By Vieta's again, <math>rst = \frac{-2008}8 = -251 \Longrightarrow -(r^3 + s^3 + t^3) = -3rst = \boxed{753}.</math>
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=== Solution 2 ===
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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>. Summing these, we have
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<math></math>\begin{align*}8\{(r + s)^3 + (s + t)^3 + (t + r)^3\} &= - 8(r^3 + s^3 + t^3)\\
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&= 1001(r + s + t) + 2008\cdot 3 = 3\cdot 2008<math>\end{align*}</math><math>, yielding the answer </math>753$.
  
 
== See also ==
 
== See also ==

Revision as of 17:56, 3 April 2008

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 $s,\ t$. Summing these, we have $$ (Error compiling LaTeX. Unknown error_msg)\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$\end{align*}$ (Error compiling LaTeX. Unknown error_msg)$, yielding the answer$753$.

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