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

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(B) Given the equation $(x - 1)^3 + 8 = 0$. This can be written as $(x - 1)^3 = -8$. Taking the cube root on both sides,we get $x - 1 = (-8)^{1/3}$. S

Vedclass · Accepted Answer

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

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(B) Given the equation $(x - 1)^3 + 8 = 0$. This can be written as $(x - 1)^3 = -8$. Taking the cube root on both sides,we get $x - 1 = (-8)^{1/3}$. Since the cube roots of unity are $1, \omega, \omega^2$,the cube roots of $-8$ are $-2, -2\omega, -2\omega^2$. Therefore,$x - 1 = -2, -2\omega, -2\omega^2$. Adding $1$ to all sides,we get $x = 1 - 2, 1 - 2\omega, 1 - 2\omega^2$. Thus,the roots are $-1, 1 - 2\omega, 1 - 2\omega^2$.

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

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