If ab + bc + ca = 0
Find the value of 1/(a2 - bc) + 1/(b2 - ca) + 1/(c2 - ab)
ab + bc + ca = 0 bc = - ab – ca ca = - ab – bc ab = -bc –ca Now, 1/(a2 - bc) + 1/(b2 - ca) + 1/(c2 - ab) = 1/(a2 - (-ab - ca)) + 1/(b2 - (-ab - bc)) + 1/(c2 - (-bc-ca)) =1/(a2 + ab+ca) + 1/(b2 + ab+bc) + 1/(c2 + bc+ca) = 1/(a(a + b + c)) + 1/(b(a + b + c)) + 1/(c(a + b + c)) = (bc + ca + ab)/(abc(a + b + c)) = 0/(abc(a + b + c)) = 0
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