# If ω$\left( { \ne 1} \right)$ is a cube root of unity and ${\left( {1 + {\omega ^2}} \right)^n} = {\left( {1 + {\omega ^4}} \right)^n}$ , then find the least positive value of n-A)$2$ B)$3$ C)$5$ D)$6$

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Hint: We can solve this using the condition of cube root of unity that,
${\omega ^3} - 1 = 0$ and use the formula $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$ to find the least positive values of n.

Since it is given that ω is cube root of unity so it satisfies the following condition,
$\Rightarrow$ ${\omega ^3} - 1 = 0$$\Rightarrow {\omega ^3} = 1$ --- (i)
On using the formula $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$, we get-
$\Rightarrow \left( {\omega - 1} \right)\left( {{\omega ^2} + \omega + 1} \right) = 0$
Now it is given that $\omega \ne 1$ so it follows that $\left( {{\omega ^2} + \omega + 1} \right) = 0$ -- (ii)
From this we can also write-
$\Rightarrow 1 + {\omega ^2} = - \omega$ --- (iii)
It is also given that ${\left( {1 + {\omega ^2}} \right)^n} = {\left( {1 + {\omega ^4}} \right)^n}$--- (iv)
Then we can write ${\omega ^4} = {\omega ^3}\omega$ and we know from equation (i) that ${\omega ^3} = 1$
So ${\omega ^4} = \omega$
Then we can write,
$\Rightarrow 1 + {\omega ^4} = 1 + \omega$
From eq. (ii) substituting the value of $1 + \omega$ , we get
$\Rightarrow 1 + {\omega ^4} = - {\omega ^2}$ --- (v)
On substituting the values of eq. (iii) and (v) in eq. (iv), we get
$\Rightarrow {\left( { - \omega } \right)^n} = {\left( { - {\omega ^2}} \right)^n}$
We have to find the least positive value of n.
So we also write the above equation as-
$\Rightarrow 1 = \dfrac{{{{\left( { - {\omega ^2}} \right)}^n}}}{{{{\left( { - \omega } \right)}^n}}}$ -- (vi)
On simplifying we get-
$\Rightarrow 1 = {\left( {\dfrac{{{\omega ^2}}}{\omega }} \right)^n}$
$\Rightarrow 1 = {\omega ^n}$ -- (vii)
Now from eq. (i) and eq. (vii), it is clear that-
$\Rightarrow {\omega ^3} = {\omega ^n} = 1$
Since the base of the raised powers is same so the powers raised will also be equal to each other which means ${a^x} = {a^y} \Rightarrow x = y$
So on applying this in the obtained equation, we get-
$\Rightarrow n = 3$
So the least positive value of n such that${\left( {1 + {\omega ^2}} \right)^n} = {\left( {1 + {\omega ^4}} \right)^n}$ is 3.

Hence, the correct answer is option ‘B’.

Note: You can also find the least positive value of n by putting value of n=$2$,$3$ in eq (vi)
For n=$2$ ,
$\Rightarrow 1 = {\left( {\dfrac{{{\omega ^2}}}{\omega }} \right)^2}$
On solving we get,
$\Rightarrow$ ${\omega ^4} = {\omega ^2}$ but we already proved that ${\omega ^4} = \omega$ and $\omega \ne {\omega ^2}$ so n=$2$ is not possible
For n=$3$ ,we get
$\Rightarrow 1 = {\left( {\dfrac{{{\omega ^2}}}{\omega }} \right)^3}$
On solving we get,
${\omega ^3} = 1$ which is given is the condition of the cube root of unity so the least positive values for n=$3$.