$a b + b c + c a = 0$ হলে $\frac{1}{a^{2} - b c} + \frac{1}{b^{2} - a c} + \frac{1}{c^{2} - a b}$ এর মান কত?

dsuc.created: 2 years ago | dsuc.updated: 1 year ago

dsuc.updated: 1 year ago

Given that, ab+bc+ca=0

$∴ a b + b c = - c a ∴ c a = - (a b + b c) = a b + c a = - b c ∴ = b c = - (a b + c a) = b c + c a = - a b ∴ = a b = - (b c + c a) A g a i n g i v e n t h a t, \frac{1}{a^{2} - b c} + \frac{1}{b^{2} - a c} + \frac{1}{c^{2} - a b} = \frac{1}{a^{2} + a b + c a} + \frac{1}{b^{2} + a b + b c} + \frac{1}{c^{2} + b c + c a} = \frac{1}{a (a + b + c)} + \frac{1}{b (b + a + c)} + \frac{1}{c (c + b + a)} = \frac{1}{a (a + b + c)} + \frac{1}{b (a + b + c)} + \frac{1}{c (a + b + c)} = \frac{1}{a (a + b + c)} \times (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) = \frac{1}{a + b + c} \times \frac{(b c + c a + a b)}{a b c} = \frac{a b + b c + c a}{a b c (a + b + c)} = \frac{0}{a b c (a + b + c)} = 0$