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

Question

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

Accepted Answer

$(x-1)^3=-8 \implies x-1=-2,\ -2\omega,\ -2\omega^2$, so $x=-1,\ 1-2\omega,\ 1-2\omega^2.$

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