Question

If $1$, $\omega$ and ${\omega }^2$ are the cube roots of unity, then $(1+\omega )(1+{\omega }^2)(1+{\omega }^4)(1+{\omega }^8)$ is equal to

• A. $1$
• B. $0$
• C. ${\omega }^2$
• D. $\omega$

Answer 1

The correct option is A

$1$

Explanation for the correct option:

Step 1: Find the values of $\mathbf{(}\mathbf{1}\mathbf{+}\mathit{\omega }\mathbf{)}\mathbf{,}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{+}{\mathit{\omega }}^{\mathbf{2}}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{+}{\mathit{\omega }}^{\mathbf{4}}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{+}{\mathit{\omega }}^{\mathbf{8}}\mathbf{)}$

The cube roots of unity are 1, $\omega =\frac{-1+\sqrt{3}i}{2}$, ${\omega }^2=\frac{-1-\sqrt{3}i}{2}$

$\begin{array}{rl}1+\omega +{\omega }^2& =1+\frac{-1+\sqrt{3}i}{2}+\frac{-1-\sqrt{3}i}{2}\end{array}$

$\begin{array}{rl}& =1+\frac{-1+\sqrt{3}i-1-\sqrt{3}i}{2}\\ & =1-1\\ & =0\end{array}$

$\Rightarrow$ $\begin{array}{rl}1+\omega & =(-\omega {)}^2\end{array}$

$\Rightarrow$$\begin{array}{rl}1+(\omega {)}^2& =-\omega \end{array}$

$\begin{array}{rl}1+(\omega {)}^4& =1+\left(\omega {)}^3\right(\omega )\\ & =1+\omega \end{array}$ $\left[\mathbf{\because }{\mathit{\omega }}^{\mathbf{3}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\right]$

$\begin{array}{rl}1+(\omega {)}^8& =1+\left(\right(\omega {)}^4{)}^2\\ & =1+(\omega {)}^2\end{array}$

Step 2: Multiply all the values to find $\mathbf{(}\mathbf{1}\mathbf{+}\mathit{\omega }\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}{\mathit{\omega }}^{\mathbf{2}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}{\mathit{\omega }}^{\mathbf{4}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}{\mathit{\omega }}^{\mathbf{8}}\mathbf{)}$

$\begin{array}{rl}\therefore (1+\omega )(1+{\omega }^2)(1+{\omega }^4)(1+{\omega }^8)& =(-\omega {)}^2(-\omega \left)\right(-\omega {)}^2(-\omega )\\ & =\left(\omega {)}^4\right(\omega {)}^2\\ & =(\omega {)}^6\\ & =\left(\right(\omega {)}^3{)}^2\\ & =1^2\\ & =\mathbf{1}\end{array}$

Hence, Option ‘A’ is Correct.