a+b+c=0,a分之1+b分之1+c分之1=负4，求a平方分之1+b平方分之1+c平方分之1=？

(1/a+1/b+1/c)^2
=1/a^2+1/b^2+1/c^2+2/ab+2/ac+2/bc
=(1/a^2+1/b^2+1/c^2)+2(1/ab+1/ac+1/bc)
=(1/a^2+1/b^2+1/c^2)+2[(a+b+c)/abc]
=(-4)^2
=16
我们知道 a+b+c=0

所以
1/a^2+1/b^2+1/c^2
=16

a+b+c=0
1/a+1/b+1/c=-4
1/aa+1/bb+1/cc
=(1/a+1/b+1/c)^2-2[1/ab+1/ac+1/bc]
=16-2(a+b+c)/abc
=16

(1/a+1/b+1/c)^2=1/(a^2)+1/(b^2)+1/(c^2)+2(1/ab+1/ac+1/bc)=
1/(a^2)+1/(b^2)+1/(c^2)+2(a+b+c)/abc =16
a+b+c=0,
所以1/(a^2)+1/(b^2)+1/(c^2)=16