What are the roots of equation-I?

Consider equation-I: z^{3}+2z^{2}+2z+1=0 and equation-II: z^{1985}+z^{100}+1=0.

  1. A. 1,\omega,\omega^{2}
  2. B. -1,\omega,\omega^{2} ✓
  3. C. 1,-\omega,\omega^{2}
  4. D. -1,-\omega,-\omega^{2}

Correct Answer: B. -1,\omega,\omega^{2}

Explanation

Equation-I can be grouped as (z^3+1) + 2z(z+1) = 0. Factoring out (z+1) gives (z+1)(z^2-z+1) + 2z(z+1) = 0, leading to (z+1)(z^2+z+1) = 0. Setting each factor to zero yields the roots z = -1 and the complex cube roots of unity, z = \omega, \omega^2.

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