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Question

If the roots of the equation (c2ab)x22(a2bc)x+(b2ac)=0 are real ane equal, show that either a=0 or (a3+b3+c3)=3abc


Solution

Given, equation is:

(c2ab)x22(a2bc)x+(b2ac)=0To prove:a=0 or a3+b3+c3=3abc

Proof: From the given equation, we have

A=(c2ab)B=2(a2bc)C=(b2ac)

It is being given that the equation has real and equal roots

D=0B24AC=0

On substituting respective values of a, b and c in the above equation, we get

[2(a2bc)]24(c2ab)(b2ac)=04(a2bc)24(c2b2ac3ab3+a2bc)=04(a4+b2c22a2bc)4(c2b2ac3ab3+a2bc)=0a4+b2c22a2bcb2c2+ac3+ab3a2bc=0a4+ab3+ac33a2bc=0a[a3+b3+c33abc]=0a=0 or a3+b3+c3=3abc


Mathematics
Secondary School Mathematics X
Standard X

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