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If the roots ff the equation (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 are equal, then prove that either a = 0 or a3 + b3 + c3 = 3abc "

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Given equation is: (c 2 – ab) x 2 – 2 (a 2 – bc) x + (b 2 – ac) = 0 To prove: a = 0 or a 3 + b 3 + c 3 = 3abc Proof: From the given equation, we have a = (c2 – ab) b = –2 (a 2 – bc) c = (b 2 – ac) It is being given that equation has real and equal roots ∴...
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Given equation is:

(c 2 – abx 2 – 2 (a 2 – bcx + (b – ac) = 0

To prove: a = 0 or a 3 + b 3 + c 3 = 3abc

Proof: From the given equation, we have

a = (c2 – ab)

b = –2 (a 2 – bc)

c = (b 2 – ac)

It is being given that equation has real and equal roots

∴ D = 0

⇒ b 2 – 4ac = 0

On substituting respective values of ab and c in above equation, we get

[–2 (a 2 – bc)]2 – 4 (c 2 – ab) (b 2 – ac) = 0

4 (a 2 – bc)2 – 4 (c 2 b 2 – ac 3 – ab 3 + a 2 bc) = 0

4 (a 4 + b 2 c 2 – 2a 2 bc) – 4 (c 2 b 2 – ac 3 – ab 3 + a 2 bc) = 0

⇒ a 4 + b 2 c 2 – 2a 2 bc – b 2 c 2 + ac 3 + ab 3 – a 2 bc = 0

⇒ a 4 + ab 3 + ac 3 –3a 2 bc = 0

⇒ a [a 3 + b 3 + c 3 – 3abc] = 0

a = 0 or a 3 + b 3 + c 3 = 3abc

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