Question

如果a+b+c=0，1/a+1/b+1/c=-4，求1/(a^2)+1/(b^2)+1/(c^2)的值

求求各位高手帮忙啊~~~~~

Answer 1

中间的合并你可以自己做，关键是有耐心观察
1/(a^2)+1/(b^2)+1/(c^2)
=(a^2b^2+b^2c^2+c^2a^2)/(a^2b^2c^2)
<<观察（1/a+1/b+1/c）^2=(a^2b^2+2ab^2c+b^2c^2+2bc^2a+c^2a^2+2ca^b)/(a^2b^2c^2)
比上面多了（2ab^2c+2bc^2a+2ca^b）/(a^2b^2c^2)=2(a+b+c)/(abc)〉〉
所以原式
=（1/a+1/b+1/c）^2-2(a+b+c)/(abc)
=16-0
=16

Answer 2

1/a^2+1/b^2+1/c^2

=(1/a+1/b+1/c)^2-2(1/(ab)-1/(ac)-1/(bc))

=(1/a+1/b+1/c)^2-2((a+b+c)/(abc))

=(1/a+1/b+1/c)^2

=16