Question

If $w$ (not equal $1$) is a cube root of unity and ${\left(1+w^2\right)}^n={\left(1+w^4\right)}^n$ , then the least positive value of $n$ is.

• A. $2$
• B. $3$
• C. $5$
• D. $6$

Answer 1

The correct option is B

$3$

Explanation for the correct answer:

Finding the least positive value of $n$:

Given, ${\left(1+w^2\right)}^n={\left(1+w^4\right)}^n$ where $w$ is the cube root of unity.

We know,

$w^3=1\mathrm{and}{w}^2+w+1=0$

So,

$$\begin{gathered} {\left(1+w^2\right)}^n={\left(1+w^4\right)}^n \\ \Rightarrow {\left(-w\right)}^n={\left(1+w\right)}^n\left[\mathbf{\because }{\mathit{w}}^{\mathbf{4}}\mathbf{=}{\mathit{w}}^{\mathbf{3}}\mathbf{.}\mathit{w}\mathbf{=}\mathit{w}\right] \\ \Rightarrow {\left(-w\right)}^n={\left(-w^2\right)}^n \\ \Rightarrow {\left(\frac{-w^2}{-w}\right)}^n=1 \\ \Rightarrow w^n=1 \end{gathered}$$

As it is given that $w$ is the cube root of unity.

$\therefore$ the least positive value of $n$ is $3$.

Hence, the correct option is B.