Question

If the relation holds true $ ab + bc + ca = 0 $, then the value of $ \dfrac{1}{{{a^2} - bc}} + \dfrac{1}{{{b^2} - ca}} + \dfrac{1}{{{c^2} - ab}} $ will be
 $
  (a){\text{ - 1}} \\
  (b){\text{ a + b + c}} \\
  (c){\text{ abc}} \\
  (d){\text{ 0}} \\
  $

Answer 1

Hint: In this problem use the given relation $ ab + bc + ca = 0 $ , to find bc in terms of ab and ac, then ca in terms of ab and bc and finally ab in terms of bc and ca. Then simplify the expression $ \dfrac{1}{{{a^2} - bc}} + \dfrac{1}{{{b^2} - ca}} + \dfrac{1}{{{c^2} - ab}} $, by using the given relation. This will help to get the right answer.

Complete step-by-step answer:
Given equation is
 $ \dfrac{1}{{{a^2} - bc}} + \dfrac{1}{{{b^2} - ca}} + \dfrac{1}{{{c^2} - ab}} $ ....................... (1)
Now it is given that $ ab + bc + ca = 0 $ ....................... (2)
So from equation (2)
bc = - (ab + ca)
Now $ \left( {{a^2} - bc} \right) $ is written as,
 $ \Rightarrow \left( {{a^2} - bc} \right) = \left( {{a^2} + ab + ca} \right) = a\left( {a + b + c} \right) $ ....................... (3)
Similarly,
 $ \Rightarrow \left( {{b^2} - ca} \right) = \left( {{b^2} + ab + bc} \right) = b\left( {a + b + c} \right) $ ....................... (4)
 $ \Rightarrow \left( {{c^2} - ab} \right) = \left( {{c^2} + bc + ca} \right) = c\left( {a + b + c} \right) $ .......................... (5)
So from equation (1) we have, $ \Rightarrow \dfrac{1}{{{a^2} - bc}} + \dfrac{1}{{{b^2} - ca}} + \dfrac{1}{{{c^2} - ab}} = \dfrac{1}{{a\left( {a + b + c} \right)}} + \dfrac{1}{{b\left( {a + b + c} \right)}} + \dfrac{1}{{c\left( {a + b + c} \right)}} $
Now simplify the above equation we have
 $ \Rightarrow \dfrac{1}{{a\left( {a + b + c} \right)}} + \dfrac{1}{{b\left( {a + b + c} \right)}} + \dfrac{1}{{c\left( {a + b + c} \right)}} = \dfrac{1}{{\left( {a + b + c} \right)}}\left[ {\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}} \right] $
 $ \Rightarrow \dfrac{1}{{\left( {a + b + c} \right)}}\left[ {\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}} \right] = \dfrac{1}{{\left( {a + b + c} \right)}}\left[ {\dfrac{{bc + ac + ab}}{{abc}}} \right] $
Now from equation (2) we have,
 $ \Rightarrow \dfrac{1}{{\left( {a + b + c} \right)}}\left[ {\dfrac{{bc + ac + ab}}{{abc}}} \right] = \dfrac{1}{{\left( {a + b + c} \right)}}\left[ {\dfrac{0}{{abc}}} \right] = 0 $
So this is the required answer.
Hence option (D) is correct.

Note – Such problems are solemnly based upon equation manipulation, we just need to use one relation to manipulate the required expression and to break it further till the point we get the least simplified form. Then simply putting the values from a given relation helps solving problems of this kind.