Q104 / 180 · COMEDK 2020 · Mathematics
If $1, \omega, \omega^{2}$ are the cube roots of unity, then $(\mathbf{i}+\omega)\left(\mathbf{i}+\omega^{2}\right)\left(\mathbf{i}+w^{4}\right)\left(\mathbf{i}+\omega^{8}\right)$ is equal io
- A0
- Correct answerB1
- C$\omega$
- D$\omega^{2}$
Solution
We have,
$$
\begin{aligned}
(1+\omega)\left(1+\omega^{2}\right)\left(1+\omega^{4}\right)\left(1+\omega^{8}\right) & \\
=&(1+\omega)\left(1+\omega^{2}\right)(1+\omega)\left(1+\omega^{2}\right) \\
& {\left[\because \omega^{4}=\omega^{3} \cdot \omega=\omega \text { and } \omega^{8}-\left(\omega^{3}\right)^{2} \cdot \omega^{2}=\omega^{3}\right] } \\
=&\left\langle(1+\omega)\left(1+\omega^{2}\right)\right)^{2}=\left(1+\omega+\omega^{2}+\omega^{3}\right)^{2} \\
=&\left(\omega^{3}\right)^{2} \quad\left[\because 1+\omega+\omega^{2}=0\right] \\
&=(1)^{2}=1 \\
{\left[\because \omega^{3}=1\right] }
\end{aligned}
$$
Hence, the correct answer is option: 'B'.