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

For a non-real cube root of unity, $\omega^3=1$ and $1+\omega+\omega^2=0$. Also, $\omega^4=\omega$. So the given condition becomes $$ (1+\omega^2)^n=(1+\omega)^n $$ Using $1+\omega=-\omega^2$ and $1+\omega^2=-\omega$, we get $$ (-\omega)^n=(-\omega^2)^n $$ Hence $$ \omega^n=\omega^{2n} $$ Therefore $$ \omega^n(1-\omega^n)=0 $$ Since $\omega\neq 0$, this gives $$ \omega^n=1 $$ Now $\omega$ is a cube root of unity different from $1$, so its order is $3$. Therefore the least positive $n$ is $$ 3 $$ Checking the options, this matches option $\text{B}$.