If ab + bc + ca = 0, then the value of `1/(a^2-bc) + 1/(b^2-ca) + 1/(c^2-ab)` will be:

Options

A. 2
B. -1
C. 0
D. 1

Answer

C

Explanation

`ab+bc+ca=0`
`ab+ca=-bc`
`rArra^(2)+ab+ca=a^(2)-bc`
`a^(2)-bc=a(a+b+c)`
Similarly
`b^(2)-ca=b(a+b+c)`
`c^(2)-ab=c(a+b+c)`
`therefore1/(a^(2)-bc)+1/(b^(2)-ac)+1/(c^(2)-ab)`
= `1/(a(a+b+c))+1/(b(a+b+c))+1/(c(a+b+c))`
= `(bc+ca+ab)/(abc(a+b+c))`
= `0/(abc.(a+b+c))=0`