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2008 AIME II Problems/Problem 7

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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.

Contents

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\cd...yielding the answer 753.

See also

2008 AIME II (Problems • 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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