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  4. If cube root of unity be 1, ω, ω2, then the roots of the equation (x-2)3+27=0 are

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Q. If cube root of unity be $1, \omega, \omega^2$, then the roots of the equation $(x-2)^3+27=0$ are

Complex Numbers and Quadratic Equations

Solution:

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