Question

If the roots of the equation (c² – ab)x² – 2(a² – bc)x + (b² – ac) = 0 are real and equal, show that either a = 0 or a³ + b³ + c³ = 3abc.

[Hint: D = 4a(a³ + b³ + c³ – 3abc). So, D = 0 ⇒ a = 0 or a³ + b³ + c³ = 3abc.]

Answer 1

Given: The quadratic (c² – ab)x² – 2(a² – bc)x + (b² – ac) = 0 has real and equal roots.

Step-wise calculation:

1. For a quadratic Ax² + Bx + C = 0, equal roots ⇒ discriminant D = B² – 4AC = 0.

2. Here A = c² – ab

B = –2(a² – bc)

C = b² – ac

3. Compute B²:

B² = 4(a² – bc)²

= 4(a⁴ + b²c² – 2a²bc)

4. Compute 4AC:

4AC = 4(c² – ab)(b² – ac)

= 4(b²c² – ac³ – ab³ + a²bc)

5. Subtract: D = B² – 4AC

= 4[a⁴ + b²c² – 2a²bc – (b²c² – ac³ – ab³ + a²bc)] 

= 4[a⁴ + ac³ + ab³ – 3a²bc] 

= 4a(a³ + b³ + c³ – 3abc)

Thus D = 4a(a³ + b³ + c³ – 3abc).

6. Since roots are equal, D = 0

⇒ 4a(a³ + b³ + c³ – 3abc ) = 0

From D = 0 we get either a = 0 or a³ + b³ + c³ = 3abc.