Algebraic equation: If ab + bc + ca = 0, find the value of...

[lPing7]

I have no clue as to how to start this problem.

If ab + bc + ca = 0, then the value of \(\displaystyle \dfrac{1}{a^2\, -\, bc}\, +\, \dfrac{1}{b^2\, -\, ca}\, +\, \dfrac{1}{c^2\, -\, ab}\) will be _____.

A) -1
B) a + b + c
C) abc
D) 0

 

[Deleted member 4993]

I have no clue as to how to start this problem.

If ab + bc + ca = 0, then the value of \(\displaystyle \dfrac{1}{a^2\, -\, bc}\, +\, \dfrac{1}{b^2\, -\, ca}\, +\, \dfrac{1}{c^2\, -\, ab}\) will be _____.

A) -1
B) a + b + c
C) abc
D) 0

Start with:

\(\displaystyle \displaystyle{\frac{1}{a^2 - bc} + \frac{1}{b^2 - ca} + \frac{1}{c^2-ba} \ = \ \frac{???}{(a^2 - bc)(b^2 - ca)(c^2-ba)}}\)

and continue....

 

[lPing7]

Start with:

\(\displaystyle \displaystyle{\frac{1}{a^2 - bc} + \frac{1}{b^2 - ca} + \frac{1}{c^2-ba} \ = \ \frac{???}{(a^2 - bc)(b^2 - ca)(c^2-ba)}}\)

and continue....

Thank you, I got how to solve it.
In the end you get ab +bc +ca by the denominator, which is 0 by denominator which = 0