Question

If $\omega (\ne 1)$ is a cube root of unity and $(1+{\omega }^2{)}^n=(1+{\omega }^4{)}^n$, then the least positive value of n is

• A. $2$
• B. $3$
• C. $5$
• D. $6$

Answer 1

The correct option is B $3$

Given:-

$\omega$ is the cube root of unity.

${(1+{\omega }^2)}^n={(1+{\omega }^4)}^n$

we know that

$1+\omega +{\omega }^2=0$ for cube root of unity. and

${\omega }^3=1$

$\Rightarrow 1+\omega =-{\omega }^2$ and

$1+{\omega }^4=1+(\omega {)}^3\omega =1+\omega$

$\therefore {(1+{\omega }^2)}^n={(1+{\omega }^4)}^n$

${(-\omega )}^n={(1+\omega )}^n$

${(-\omega )}^n={(-{\omega }^4)}^n$

${\left(\frac{\omega }{{\omega }^2}\right)}^n=1$

${\left(\frac{1}{\omega }\right)}^n=1$

${\left(\omega \right)}^n=1={\omega }^3$

We also know that ${\omega }^3=1$

$\therefore n=3$

Hence, the answer is $3$.