Answer :

Given that the roots of the equation (c2 – ab)x2 – 2(a2 – bc)x + (b2 – ac) = 0 are real and equal

Comparing with standard quadratic equation ax2 + bx + c = 0


a = (c2 – ab) b = – 2(a2 – bc) c = (b2 – ac)


Thus D = 0


Discriminant D = b2 – 4ac = 0


[ – 2(a2 – bc)]2 – 4(c2 – ab) (b2 – ac) = 0


4(a4 – 2a2bc + b2c2) – 4(b2c2 – ac3 – ab3 + a2bc) = 0


using (a – b)2 = a2 – 2ab + b2


a4 – 2a2bc + b2c2 – b2c2 + ac3 + ab3 – a2bc = 0


a4 – 3a2bc + ac3 + ab3 = 0


a (a3 – 3abc + c3 + b3) = 0


a = 0 or (a3 – 3abc + c3 + b3) = 0


Hence proved a = 0 or a3 + c3 + b3 = 3abc


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