If the cube roots of unity are 1, omega, omeg | vedclass

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(D) Given the equation $(x - 2)^3 + 27 = 0$.This can be rewritten as $(x - 2)^3 = -27$.Taking the cube root on both sides,we get $x - 2 = (-27)^{1/3}$

Vedclass · Accepted Answer

$ - 1, 2 - 3\omega, 2 - 3\omega^2$

Vedclass · Answer

(D) Given the equation $(x - 2)^3 + 27 = 0$.This can be rewritten as $(x - 2)^3 = -27$.Taking the cube root on both sides,we get $x - 2 = (-27)^{1/3}$.Since $(-27)^{1/3} = -3(1)^{1/3}$,and the cube roots of unity are $1, \omega, \omega^2$,we have:$x - 2 = -3(1), -3(\omega), -3(\omega^2)$.$x - 2 = -3, -3\omega, -3\omega^2$.Adding $2$ to each term,we get:$x = 2 - 3, 2 - 3\omega, 2 - 3\omega^2$.$x = -1, 2 - 3\omega, 2 - 3\omega^2$.

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

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