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

Question

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

Dr. Abubacker · Accepted Answer

(b)
We have, $\left( 1 + \omega^{2} \right)^{n} = \left( 1 + \omega^{4} \right)^{n}$
$\Rightarrow \ \ \ \ ( - \omega)^{n} = \left( - \omega^{2} \right)^{n}$
$\Rightarrow \ \ \ \omega^{n} = 1$
$\Rightarrow \ \ \ \ \ \ n = 3$ is least positive value of $n$

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

Explanation

Reviewed by Dr. Abubacker

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