Question

If ab+bc+ca =0, then find the value of :
1/(a² – bc) + 1/(b² – ca) + 1/(c² – ab)

Answer 1

1 / (a² - bc) +  1 / (b² - ca) +  1 / (c² - ab) = 0

• Given, ab + bc + ca = 0

or, ab = - bc - ca  - (i)

or, bc = - ab - ca  - (ii)

or, ca = - ab - bc  - (iii)

• Therefore, 1 / (a² - bc) = 1 / {a² - (-ab - ca)}  from (i)

or,  1 / (a² - bc) = 1 / (a² + ab + ca)

or,  1 / (a² - bc) = 1 / { a (a + b + c) }  - (iv)

• Similarly,  1 / (b² - ca) = 1 / {b² - (- ab - bc)}  from (ii)

or,  1 / (b² - ca) = 1 / (b² + ab + bc)

or,  1 / (b² - ca) = 1 / { b (b + a + c) }  - (v)

• Similarly 1 / (c² - ab) = 1 / {c² - (-bc - ca)}  from (iii)

or,  1 / (c² - ab) = 1 / (c² + bc + ca)

or, 1 / (c² - ab) = 1 / { c (c + b + a) } - (vi)

• Now, adding (iv), (v), and (vi), we get,

1 / { a (a + b + c) }  + 1 / { b (b + a + c) }  + 1 / { c (c + b + a) }  

or, (bc + ac + ab) / {abc (a + b + c)}

or, 0 / {abc (a + b + c)}   [ Given, ab + bc + ca = 0]

or, 0

• Therefore,

1 / (a² - bc) +  1 / (b² - ca) +  1 / (c² - ab) = 0