已知,a,b是方程x2+px+1=0的两根；c,d方程x2+qx+1=0的两根；
可得：a+b = -p ,ab = 1 ；c+d = -q ,cd = 1 .
(a-c)(b-c)(a+d)(b+d)
= [ab+c^2-c(a+b)]·[ab+d^2+d(a+b)]
= [cd+c^2-c(a+b)]·[cd+d^2+d(a+b)]
= [c(d+c-(a+b)]·[d(c+d+a+b)]
= cd·[(c+d)-(a+b)]·[(c+d)+(a+b)]
= (c+d)^2-(a+b)^2
= q^2-p^2