Problem: The fraction a2+c2βb2+2aca2+b2βc2+2abβ is (with suitable restrictions on the values of a,b, and c):
Answer Choices:
A. irreducible
B. reducible to -1
C. reducible to a polynomial of three terms
D. reducible to a+bβcaβb+cβ
E. reducible to aβb+ca+bβcβ
Solution:
a2+c2βb2+2aca2+b2βc2+2abβ=(a+c)2βb2(a+b)2βc2β=(a+c+b)(a+cβb)(a+b+c)(a+bβc)β
=a+cβba+bβcβ with (a+c)2ξ =b2