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

Revision as of 18:01, 28 May 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 $(r+s+t)^3 = r^3 + s^3 + t^3 + 3r^2s + 3s^2r + 3s^2t + 3t^2s + 3r^2t + 3t^2r + 6rst$ 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 $a^3+b^3+c^3 = 3abc = 3(r+s)(s+t)(t+r) = 3(-t)(-r)(-s) = 3[-(-251)] = \boxed{753}.$

@@ Line 31: / Line 31: @@
 <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>
+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 + 6rst</math>
 Substituting:
 <cmath>(r+s+t)^3 - 6rst + r^3+s^3+t^3 = (r + s)^3 + (s + t)^3 + (t + r)^3</cmath>

2008 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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