Q13/105)if ab + bc +ca=0,then find 1/(a^2-bc) +1(/b^2-ca) +(1/c^2-ab)

ab + bc + ca = 0

so bc = - ab - ca = -a(b+c)

so a^2 - bc = a(a + b+ c)

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

similarly

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

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

adding we get (1/a^2-bc) +(1/b^2-ca) +(1/c^2-ab)= 1/(a+b+c) ( 1/a + 1/b+ 1/c)

= ( 1/a + 1/b+ 1/c)/(a+b+c)
=(bc + ac + ab)/((a+b+c)(abc)) = 0