Question

If $1,\omega and{\omega }^2$are the cube roots of unity, ${\left(3+{\omega }^2+{\omega }^4\right)}^6$ is equal to

• A. $64$
• B. $729$
• C. $2$
• D. $0$

Answer 1

The correct option is A

$64$

Explanation of the correct option:

Step1. Define the cube root of unity.

Given,

$1,\omega and{\omega }^2$are cube root of unity.

$$\begin{gathered} 1+\omega +{\omega }^2=0\_\_\_\_\left(1\right)\left[\mathbf{\because }\mathbf{From}\mathbf{}\mathbf{definition}\mathbf{,}\mathbf{\omega }\mathbf{=}\frac{\mathbf{-}\mathbf{1}\mathbf{+}\mathbf{i}\sqrt{\mathbf{3}}}{\mathbf{2}}\mathbf{}\mathbf{,}{\mathbf{\omega }}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{-}\mathbf{1}\mathbf{-}\mathbf{i}\sqrt{\mathbf{3}}}{\mathbf{2}}\right] \\ 1\times \omega \times {\omega }^2=1\_\_\_\_\left(2\right) \end{gathered}$$

Step2. Find the value of ${\left(3+{\omega }^2+{\omega }^4\right)}^{\mathbf{6}}$:

$$\begin{gathered} {\left(3+{\omega }^2+{\omega }^4\right)}^6 \\ ={\left(3+{\omega }^2+\left({\omega }^3\right)\left(\omega \right)\right)}^6 \\ ={\left(3+{\omega }^2+\omega \right)}^6\mathbf{\left[}\mathbf{\because }\mathbf{}\mathbf{From}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{}\mathbf{,}{\mathbf{\omega }}^{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{\right]} \\ ={\left(2+1+{\omega }^2+\omega \right)}^6 \\ ={\left(2+0\right)}^6\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{From}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{,}\mathbf{1}\mathbf{+}\mathbf{\omega }\mathbf{+}{\mathbf{\omega }}^{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{]} \\ =2^6 \\ =64 \end{gathered}$$

Hence, Option(A) is the correct answer.