If $\omega = \frac{- 1 + \sqrt{3}i}{2}$, then $\left( 3 + \omega + 3\omega^{2} \right)^{4}$ is

Question

Dr. Abubacker · Accepted Answer

(c)
$\left( 3 + \omega + 3\omega^{2} \right)^{4} = \left\lbrack 3 + \left( 1 + \omega^{2} \right) + \omega \right\rbrack^{4}$
$= \lbrack - 3\omega + \omega\rbrack^{4}$
$= ( - 2\omega)^{4}$
$= 16\omega$

Q. If $\omega = \frac{- 1 + \sqrt{3}i}{2}$, then $\left( 3 + \omega + 3\omega^{2} \right)^{4}$ is

Explanation

Reviewed by Dr. Abubacker

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