If $\omega$ is a complex cube root of unity, find the value of $(1+\omega )(1+{\omega }^2)(1+{\omega }^4)(1+{\omega }^8)$.

Solution

1. Given that ${\omega }^3=1$ and $1+\omega +{\omega }^2=0$, we simplify the terms: $1+\omega =-{\omega }^2$, $1+{\omega }^2=-\omega$, $$\begin{array}{rl}1+{\omega }^4& =1+\omega \\ & =-{\omega }^2\end{array}$$, and $$\begin{array}{rl}1+{\omega }^8& =1+{\omega }^2\\ & =-\omega \end{array}$$.
2. Substitute these values into the original expression: $$\begin{array}{rl}(-{\omega }^2)(-\omega )(-{\omega }^2)(-\omega )& ={\omega }^2\cdot \omega \cdot {\omega }^2\cdot \omega \\ & ={\omega }^6\end{array}$$.
3. Since ${\omega }^3=1$, it follows that $$\begin{array}{rl}{\omega }^6& =({\omega }^3{)}^2\\ & =1^2\\ & =1\end{array}$$.
4. Hence the answer is (B).