If 1, omega, omega^2 are the cube roots of un | vedclass

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(B) Given equation: $8z^3 - 12z^2 + 6z - 28 = 0$ ...$(i)$By inspection,$z = 2$ is a root because $8(8) - 12(4) + 6(2) - 28 = 64 - 48 + 12 - 28 = 0$.Di

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$2, \frac{3\omega + 1}{2}, \frac{3\omega^2 + 1}{2}$

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(B) Given equation: $8z^3 - 12z^2 + 6z - 28 = 0$ ...$(i)$By inspection,$z = 2$ is a root because $8(8) - 12(4) + 6(2) - 28 = 64 - 48 + 12 - 28 = 0$.Dividing the equation by $(z - 2)$,we get:$8z^3 - 16z^2 + 4z^2 - 8z + 14z - 28 = 0$$8z^2(z - 2) + 4z(z - 2) + 14(z - 2) = 0$$(z - 2)(8z^2 + 4z + 14) = 0$$(z - 2)(4z^2 + 2z + 7) = 0$For $4z^2 + 2z + 7 = 0$,the roots are $z = \frac{-2 \pm \sqrt{4 - 112}}{8} = \frac{-2 \pm \sqrt{-108}}{8} = \frac{-2 \pm 6\sqrt{3}i}{8} = \frac{-1 \pm 3\sqrt{3}i}{4}$.We know $\omega = \frac{-1 + i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$.Consider $\frac{3\omega + 1}{2} = \frac{3(\frac{-1 + i\sqrt{3}}{2}) + 1}{2} = \frac{-3 + 3i\sqrt{3} + 2}{4} = \frac{-1 + 3i\sqrt{3}}{4}$.Similarly,$\frac{3\omega^2 + 1}{2} = \frac{3(\frac{-1 - i\sqrt{3}}{2}) + 1}{2} = \frac{-3 - 3i\sqrt{3} + 2}{4} = \frac{-1 - 3i\sqrt{3}}{4}$.Thus,the roots are $2, \frac{3\omega + 1}{2}, \frac{3\omega^2 + 1}{2}$.

If $1, \omega, \omega^2$ are the cube roots of unity,then the roots of the equation $8z^3 - 12z^2 + 6z - 28 = 0$ are

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