因为a+b+c=0所以c=-(a+b)
如果是a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3的话
=a[1/b-1/(a+b)]+b[-1/(a+b)+1/a)-(a+b)(1/a+1/b)+3
=a(a+b-b)/[(a+b)b]+b(-a+a+b)/[a(a+b)]-(a+b)²/(ab)+3
=a²/[(a+b)b]+b²/[a(a+b)]-(a+b)²/(ab)+3
=[a³+b³-(a+b)³]/[ab(a+b)]+3
=[a³+b³-(a³+3a²b+3ab²+b³)]/[ab(a+b)]+3
=-(3a²b+3ab²)/[ab(a+b)]+3
=-[3ab(a+b)]/[ab(a+b)]+3
=-3+3
=0