2008 AIME II Problems/Problem 7

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

2008 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2008 AIME II Problems/Problem 7

Problem

Contents

Solution

Solution 1

Solution 2

See also