If the roots of the equation (c2 – ab)x2 – 2(a2 – bc)x + (b2 –

1.	If the roots of the equation (c2 – ab)x2 – 2(a2 – bc)x + (b2 – ac) = 0 for a ≠ 0 are real and equal, then the value of a3 + b3 + c3 is :(a) abc (b) 3abc (c) 0 (d) None of these
Answer» (b) 3abc Given that the roots are real and equal, D =0 ⇒ b² – 4ac = 0 for ax² + bx + c = 0. ∴ [–2(a² – bc)]² – 4(c² – ab) (b² – ac) = 0 ⇒ 4[a⁴ + b²c² – 2a²bc – c²b² + ac³ + ab³ – a²bc) = 0 ⇒ 4a (a³ + b³ + c³ – 3 abc) = 0 ⇒ a³ + b³ + c³ = 3abc.

If the roots of the equation (c2 – ab)x2 – 2(a2 – bc)x + (b2 – ac) = 0 for a ≠ 0 are real and equal, then the value of a3 + b3 + c3 is :(a) abc (b) 3abc (c) 0 (d) None of these

Answer»

(b) 3abc

Given that the roots are real and equal,

D =0 ⇒ b² – 4ac = 0 for ax² + bx + c = 0.

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

⇒ 4[a⁴ + b²c² – 2a²bc – c²b² + ac³ + ab³ – a²bc) = 0

⇒ 4a (a³ + b³ + c³ – 3 abc) = 0

⇒ a³ + b³ + c³ = 3abc.