Q45 / 120 · VITEEE 2019 · Mathematics
If $\omega=\frac{-1+\sqrt{3} i}{2}$ then $\left(3+\omega+3 \omega^{2}\right)^{4}$ is
- A16
- B$-16$
- Correct answerC$16 \omega$
- D$16 \omega^{2}$
Solution
Let $\omega=\frac{-1+\sqrt{3} i}{2}$
Now, $1+\omega+\omega^{2}=0$, and $\omega^{3}=1$
then $\left(3+\omega+3 \omega^{2}\right)^{4}=\left(3+\omega+3 \omega^{2}+2 \omega-2 \omega\right)^{4}$
$$
\begin{array}{l}
=\left(3+3 \omega+3 \omega^{2}-2 \omega\right)^{4}=\left[3\left(1+2+\omega^{2}\right)-2 \omega\right]^{4} \\
=(0-2 \omega)^{4}=16 \omega^{4}=16 \omega \quad\left(\because \omega^{3}=1\right)
\end{array}
$$
Hence, the correct answer is option: 'C'.