Q. If the cube root of unity are $1,\ \omega,\ \omega^{2}$, then the roots of the equation $(x - 1)^{3} + 8 = 0$, are
A) $- 1,\ 1 + 2\omega,\ 1 + 2\omega^{2}$
B) $- 1,\ 1 - 2\omega,\ 1 - 2\omega^{2}$
C) $- 1,\ - 1,\ - 1$
D) $- 1,\ - 1 + 2\omega,\ - 1 - 2\omega^{2}$
Explanation
(b)
Given, $(x - 1)^{3} = ( - 2)^{3}\ \ \Rightarrow \ \ \ \left( \frac{x - 1}{- 2} \right) = (1)^{1/3}$
$\therefore$ Cube roots of $\left( \frac{x - 1}{- 2} \right)$ are $1,\ \omega$ and $\omega^{2}$
$\Rightarrow$ Cube roots of $(x - 1)$ are $- 2,\ - 2\omega$ and $- 2\omega^{2}$
$\Rightarrow \$ Cube roots of $x$ are $- 1,\ 1 - 2\omega$ and $1 - 2\omega^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.