If the cube roots of unity are $1, \omega, \omega^2$ then the roots of the equation $(x-1)^3+8=0$, are

If the cube roots of unity are $1, \omega, \omega^2$ then the roots of the equation $(x-1)^3+8=0$, are
  1. $-1,-1+2 \omega,-1-2 \omega^2$
  2. $-1,-1,-1$
  3. $-1,1-2 \omega, 1-2 \omega^2$
  4. $-1,1+2 \omega, 1+2 \omega^2$

Solution

$(x-1)^3+8=0 \Rightarrow(x-1)=(-2)(1)^{1 / 3}$ $\Rightarrow x-1=-2$ or $-2 \omega$ or $-2 \omega^2$ or $n=-1$ or $1-2 \omega$ or $1-2 \omega^2$.

Asked in: JEE Main 2005

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