Question

If the roots of the equation (csq-ab) xsq-2(asq-bc)x+bsq-ac=0 are equal prove that a=0 /a cube +b cube+c cube=3abc

Answer 1

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