Question

If the roots of the equation (c² – ab)x² – 2(a² – bc)x + (b² – ac) = 0 are real are equal, show that either 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 (discriminant = 0).

Step-wise calculation:

1. Discriminant D = [–2(a² – bc)]² – 4(c² – ab)(b² – ac)

2. So, `D/4 = (a^2 - bc)^2 - (c^2 - ab)(b^2 - ac) = 0`.

3. Expand: (a² – bc)² = a⁴ – 2a²bc + b²c². 

(c² – ab)(b² – ac) = b²c² – ac³ – ab³ + a²bc.

4. Subtracting gives 0 = a⁴ – 2a²bc + b²c² – (b²c² – ac³ – ab³ + a²bc)

= a⁴ – 3a²bc + ac³ + ab³

5. Factor out a: 0 = a(a³ + b³ + c³ – 3abc).

Therefore, either a = 0 or a³ + b³ + c³ = 3abc.