Q. If the cube roots of unity are $1,\ \omega,\ \omega^{2}$, then the roots of the equation $(x - 2)^{3} + 27 = 0$ are
A) -1, -1, -1
B) $- 1,\ - \omega,\ - \omega^{2}$
C) $- 1,\ 2 + 3\omega,\ 2 + 3\omega^{2}$
D) $- 1,\ 2 - 3\omega,\ 2 - 3\omega^{2}$
Explanation
(d)
Given, $(x - 2)^{3} = - 27 = - 3^{3}$
$\Rightarrow \ \ \ (x - 2) = - 3(1)^{1/3}$
$\Rightarrow \ \ \ (x - 2) = - 3,\ - 3\omega,\ - 3\omega^{2}$
$\Rightarrow \ \ \ \ x = - 1,\ 2 - 3\omega,\ 2 - 3\omega^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.