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Question

(C) Let $x = (27)^{1/3}$.Then $x^3 = 27$,which implies $x^3 - 27 = 0$.This can be factored as $(x - 3)(x^2 + 3x + 9) = 0$.The roots are $x = 3$ and th

Vedclass · Accepted Answer

$3, 3\omega, 3\omega^2$

Vedclass · Answer

(C) Let $x = (27)^{1/3}$.Then $x^3 = 27$,which implies $x^3 - 27 = 0$.This can be factored as $(x - 3)(x^2 + 3x + 9) = 0$.The roots are $x = 3$ and the roots of $x^2 + 3x + 9 = 0$.Using the quadratic formula,$x = \frac{-3 \pm \sqrt{9 - 36}}{2} = \frac{-3 \pm i\sqrt{27}}{2} = 3 \left( \frac{-1 \pm i\sqrt{3}}{2} \right)$.Since $\omega = \frac{-1 + i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$,the roots are $3, 3\omega, 3\omega^2$.

$(27)^{1/3} = $

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