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

1
by shruti7

2015-10-05T17:09:52+05:30

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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