Question

If a+b+c=0 then find 1/-a2+b2+c2 + 1/a2-b2+c2 + 1/a2+b2-c2

Answer 1

GIVEN THAT a+b+c =0 -------1²

now, 1/a²+b²+c² + 1/b²+c²-a²  + 1/c²+a²-b²

=1/a²+(b+c)(b-c)  + 1/b²+(c+a)(c-a) + 1/c²+(c+b)(a-b)

=1/a²+(-a)(b-c) + 1/b²+ (-b)(c-a) + 1/c²+ (-c)(a-b)    (from 1)

=1/a(a-b+c) + 1/b(a+b-c) = 1/c(c-a+b)

=1/a(a+c-b) + 1/b(a+b-c) + 1/c(b+c-a)

=1/a(-b-b) + 1/b(-c-c) + 1/ c (-a-a)                             (frpom 1)

=1/-2ab + 1/-2bc + 1/-2ac 

= 1/-2 (1/ab + 1/bc+ 1/ac )

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

=-1/2 (0)  (from 1)

= 0

Answer 2

a+b+c=0

or,a+b=(-c)

or,(a+b)²=(-c)² [squaring both sides]

or, a²+b²+2ab=c²

or,a²+b²-c²=(-2ab)

or, 1/a²+b²-c²=(-1/2ab)

as a same way,

1/b²+c²-a²=(-1/2bc) and, 1/c²+a²-b²=(-2bc)

finally, put their value

.

. . (-1/2bc)+(-1/2ca)+(-1/2ab)

=(-1/2bc-1/2ac-1/2ab)

=(-a-b-c/2abc)

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

=0 [ cause, a+b+c=0]

so, the answer is 0 .