If \omega\neq1 is a cube root of unity, then what are the solutions of (z-100)^{3}+1000=0?

  1. A. 10(1-\omega), 10(10-\omega^{2}) 100
  2. B. 10(10-\omega), 10(10-\omega^{2}), 90 ✓
  3. C. 10(1-\omega), 10(10-\omega^{2}), 1000
  4. D. (1+\omega) (10+\omega^{2}), −1

Correct Answer: B. 10(10-\omega), 10(10-\omega^{2}), 90

Explanation

The equation is (z-100)^3 = -1000, which can be written as (\frac{z-100}{-10})^3 = 1. The cube roots of unity are 1, \omega, \omega^2. Thus, \frac{z-100}{-10} = 1, \omega, \omega^2. Solving for z gives z = 100 - 10 = 90, z = 100 - 10\omega = 10(10-\omega), and z = 100 - 10\omega^2 = 10(10-\omega^2).

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