Question

if ab+bc+ca=0, find the value of (1/a²-bc)+(1/b²-ca)+(1/c²-ab)

Answer 1

Given  ab + bc + ca = 0
 =>  bc = - ab - ca = -a (b+c)
=>  a² - bc  =  a (a - b - c)

similarly,  b² - ca = b (b - c - a)          and    c² - ab = c (c - a - b)

==========
1/(a²-bc)  + 1/(b²-ca)  + 1/(c² -ab)
   = [(b²-ca)(c²-ab) + (c²-ab)(a²-bc) + (a²-bc)(b²-ca) ] / [(a²-bc)(b²-ca)(c²-ab)]

we simplify the numerator.
   = [ b²c² - ac³ - a b³ +a²bc +a²c² - ba³ - bc³ + ab²c + a²b² - cb³ -ca³ + abc² ]
   = [ b²(c²-ab+ac+a²-bc) + c²(-ac+a²-bc+ab) + a² (bc-ab-ca)

  we use the given identity.

  = [ b² (c² +a²+2 ac) + c² (a² +2ab) + a²(2bc) ]
  = [ b² (c + a)² +  a² c² + 2a²bc + 2abc² ]
  = [ { bc + ba}² + a²c² + 2ac (ab + bc) ]

we use the given identity again.

  = [  {-ac}²  + a² c² + 2 ac (-ac) ]
  = 0

so the answer is 0.

Answer 2

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) abc
= (a+b+c) abc/(a+b+c) abc
= 0