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

Revision as of 16:56, 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$.

@@ Line 6: / Line 6: @@
 Find <math>(r + s)^3 + (s + t)^3 + (t + r)^3</math>.
+__TOC__
 == 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
+=== Solution 1 ===
+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>
-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>
+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>
+=== Solution 2 ===
+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
+<math></math>\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<math>\end{align*}</math><math>, yielding the answer </math>753$.
 == 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
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