可设x³+3x+2=
= (x-a)(x-b)(x-c)
= x³-(a+b+c)x²+(ab+bc+ac)x-abc
则对号入座得 a+b+c=0 ab+bc+ac=3 abc=-2
则以(a+1/a),(b+1/b) ,(c+1/c)为根的多项式
可以表示为x³-(a+b+c+1/a+1/b+1/c)x²+[(a+1/a)(b+1/b)+ (b+1/b)(c+1/c)+(c+1/c)(a+1/a)]x-(a+1/a)(b+1/b)(c+1/c)
a+b+c+1/a+1/b+1/c=0+(ab+bc+ac)/abc=-3/2
(a+1/a)(b+1/b)+(b+1/b)(c+1/c)+(c+1/c)(a+1/a)
=ab+bc+ac+(a+b+c)/abc+ a/b+b/a+a/c+c/a+b/c+c/b
(a+b+c)²=a²+b²+c²+2(ab+bc+ac)=a²+b²+c²+6=0 a²+b²+c²=-6(a,b,c是复数?)
楼主,是不是题目错了?确定是x^3+3x+2=0由根a,b,