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 $(r + s)^3 + (s + t)^3 + (t + r)^3$.

Solution

By Vieta’s sums, we have $r+s+t = 0$, and so the desired answer is $(r+s)^3 + (s+t)^3 + (t+r)^3 = -(r^3 + s^3 + t^3)$. Additionally, using the factorization \[r^3 + s^3 + t^3 - 3rst = (r+s+t)(r^2 + s^2 + t^2 - rs - st - tr) = 0\] Thus $r^3 + s^3 + t^3 = 3rst$. By Vieta’s again, we have $rst = \frac{-2008}8 = -251 \Longrightarrow -(r^3 + s^3 + t^3) = -3rst = \boxed{753}.$

See also

2008 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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All AIME Problems and Solutions
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