2008 AIME II Problem 7

Attempt Problem 7 of the 2008 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AIME II solutions, or check the answer key.

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7.

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.

Answer

Answer: 753

Concepts:Vieta’s Formulas polynomial sum and difference of cubes

Difficulty rating: 2410

Small hint:

There is no

x^2

term, so

r + s + t = 0

and each of

r + s,

s + t,

t + r

is the negative of a root

Big hint:

When

x + y + z = 0,

the identity

x^3 + y^3 + z^3 = 3xyz

applies; Vieta gives the product of the roots from the constant term

Solution:

The cubic has no

x^2

term, so

r + s + t = 0

by Vieta’s formulas. Hence

r + s = -t,

s + t = -r,

and

t + r = -s,

and the desired sum is

\begin{aligned} &(-t)^3 + (-r)^3 \\ &\quad {}+ (-s)^3 \\ &= -(r^3 + s^3 + t^3). \end{aligned}

Whenever

r + s + t = 0,

the identity

r^3 + s^3 + t^3 - 3rst

= (r + s + t)

(r^2 + s^2 + t^2 - rs - st - tr)

gives

r^3 + s^3 + t^3 = 3rst.

By Vieta’s formulas,

rst = -\frac{2008}{8} = -251,

so

r^3 + s^3 + t^3 = -753,

and the answer is

-(-753) = 753.