Problem:
Let a,b,c be the roots of x3+kx+1. What is
a3b2+a2b3+b3c2+b2c3+c3a2+c2a3?
Answer Choices:
A. −k
B. −k+1
C. 1
D. k−1
E. k
Solution:
We have a+b+c=0, abc=−1 and ab+bc+ac=k from Vieta’s Formulas. Factoring adjacent terms, we get the expression equals:
a2b2(a+b)+b2c2(b+c)+a2c2(a+c)
=a2b2(−c)+b2c2(−a)+a2c2(−b)
=−abc(ab+bc+ac)=(E) k
Remark. The full factorization of the given sum is
a3b2+a2b3+b3c2+b2c3+c3a2+c2a3=(ab+bc+ca)2(a+b+c)−2abc(a+b+c)2−abc(ab+bc+ca)
In the given problem, the first two terms cancel out as a+b+c=0.
The problems on this page are the property of the MAA's American Mathematics Competitions