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Asked by Shantham Last Modified
Answer 
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
∴ D = 0
⇒ b 2 – 4ac = 0
On substituting respective values of a, b 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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