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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) $-1,1+2 \omega, 1+2 \omega^{2}$
in Mathematics by Diamond (74,866 points) | 10 views

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Answer

(3)

Explanation

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

by Diamond (74,866 points)

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