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If $\omega=\frac{-1+\sqrt{3} i}{2}$ then $\left(3+\omega+3 \omega^{2}\right)^{4}$ is
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Verified Answer
The correct answer is:
$16 \omega$
Let $\omega=\frac{-1+\sqrt{3} i}{2}$
Now, $1+\omega+\omega^{2}=0$, and $\omega^{3}=1$
then $\left(3+\omega+3 \omega^{2}\right)^{4}=\left(3+\omega+3 \omega^{2}+2 \omega-2 \omega\right)^{4}$
$$
\begin{array}{l}
=\left(3+3 \omega+3 \omega^{2}-2 \omega\right)^{4}=\left[3\left(1+2+\omega^{2}\right)-2 \omega\right]^{4} \\
=(0-2 \omega)^{4}=16 \omega^{4}=16 \omega \quad\left(\because \omega^{3}=1\right)
\end{array}
$$
Now, $1+\omega+\omega^{2}=0$, and $\omega^{3}=1$
then $\left(3+\omega+3 \omega^{2}\right)^{4}=\left(3+\omega+3 \omega^{2}+2 \omega-2 \omega\right)^{4}$
$$
\begin{array}{l}
=\left(3+3 \omega+3 \omega^{2}-2 \omega\right)^{4}=\left[3\left(1+2+\omega^{2}\right)-2 \omega\right]^{4} \\
=(0-2 \omega)^{4}=16 \omega^{4}=16 \omega \quad\left(\because \omega^{3}=1\right)
\end{array}
$$
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