Question

(a-b) (a+b)+(b-c) (b+c)+(c-a) (c+a)=0​

Answer 1

see my bio

_____________LEGENDARYBOY____________

Answer 2

Answer:

ab+bc+ca=0

ab+bc = -ca

Or ab+ca= -bc

Or bc+ca = -ab

Now

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

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

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

= 1/(a+b+c) * (bc + ca + ab)/abc

= (ab+bc+ca)/(a+b+c) ab

= (a+b+c) abc/(a+b+c) ABC

I HOPE IT HELPED YOU IF YES THEN PLEASE MARK ME AS BRANIST, FOLLOW ME AND IF YOU WANT GIVE ME A THANKS. :) IT REALLY APPRECIATING MY FRIEND