Multiple choice

The equation $(x-a)^{3}+(x-b)^{3}+(x-c)^{3}=0$ has

  1. all the roots are real

  2. one real and two imaginary
  3. 3 real roots namely $x=a,\ x=b,\ x=c$

  4. no real roots
Reveal answer Fill a bubble to check yourself
B Correct answer
AI explanation

Let f(x) equal (x-a)^3 + (x-b)^3 + (x-c)^3. Taking the derivative gives f'(x) equal to 3(x-a)^2 + 3(x-b)^2 + 3(x-c)^2, which is strictly positive for all real x unless a = b = c. Since the function is strictly increasing everywhere, it crosses the x-axis exactly once. Therefore, the cubic equation has exactly one real root and two complex imaginary roots.

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