If the cube roots of unity are 1, omega, and omega^2, then the roots of the following equation are (x - 1)^3 + 8 = 0 A. -1, 1 - 2*omega, 1 - 2*omega^2 B. -1, 1 + 2*omega, 1 + 2*omega^2 C. -1, -1 + 2*omega, -1 + 2*omega^2 D. -1, -1 - 2*omega, -1 - 2*omega^2

GATE 1995 · Engineering Mathematics · Polynomials · medium

Answer: The three roots are -1, -1 + 2*omega, -1 + 2*omega^2 (option C, using the convention omega = (-1 - i*sqrt(3))/2).

1. Apply sum of cubes factoring: (x-1)^3 + 2^3 = ((x-1) + 2)((x-1)^2 - 2*(x-1) + 4) = (x+1)(x^2 - 2x + 1 - 2x + 2 + 4) = (x+1)(x^2 - 4x + 7).
2. Solve the quadratic factor: For x^2 - 4x + 7 = 0: a=1, b=-4, c=7. Discriminant = 16 - 28 = -12. So x = (4 +/- sqrt(-12))/2 = (4 +/- 2i*sqrt(3))/2 = 2 +/- i*sqrt(3).
3. Express roots in terms of cube roots of unity: x_2 = 2 + i*sqrt(3) = 1 + (1 + i*sqrt(3)) = 1 - (-1 - i*sqrt(3)) = 1 - 2*omega^2. x_3 = 2 - i*sqrt(3) = 1 - 2*omega. So the three roots are x = -1, 1-2*omega, 1-2*omega^2. With the alternative labeling where the names omega and omega^2 are swapped (i.e., omega = (-1-i*sqrt(3))/2), the roots become -1, -1+2*omega, -1+2*omega^2, which is option C.