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

Question

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

Tardigrade Admin · Accepted Answer

(x-2)3=(-3)3 x-2=+3(-1)1 / 3 x-2=-3,-3 ω,-3 ω2 x=-1,2-3 ω, 2-3 ω2

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

$- 1, 3 - 2 ω, 3 - 2 ω^{2}$

$- 1, 1 - 2 ω, 1 - 2 ω^{2}$

$- 1, 1 + 2 ω, 1 + 2 ω^{2}$

$- 1, 2 - 3 ω, 2 - 3 ω^{2}$

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

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

−1,3−2ω,3−2ω2

−1,1−2ω,1−2ω2

−1,1+2ω,1+2ω2

−1,2−3ω,2−3ω2