Problem:
Let r,s, and t be the three roots of the equation
8x3+1001x+2008=0
Find (r+s)3+(s+t)3+(t+r)3.
Solution:
Because the equation is cubic and there is no x2 term, the sum of the roots is 0; that is, r+s+t=0. Therefore,
(r+s)3+(s+t)3+(t+r)3=(−t)3+(−r)3+(−s)3=−(r3+s3+t3).
Because r is a root, 8r3+1001r+2008=0, and similarly for s and t. Therefore, 8(r3+s3+t3)+1001(r+s+t)+3⋅2008=0, and
r3+s3+t3=−81001(r+s+t)+3⋅2008=−83⋅2008=−753
and hence (r+s)3+(s+t)3+(t+r)3=−(r3+s3+t3)=753.
The problems on this page are the property of the MAA's American Mathematics Competitions