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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)

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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.

=> 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.