If $\omega ( \ne 1)$ is a complex cube root of unity and ${\left( {1 + {\omega ^4}} \right)^n} = {\left( {1 + {\omega ^8}} \right)^n}$, then the least positive integer value of n is ?
A. 2
B. 3
C. 6
D. 12

Hint: Here we will use the properties of cube roots i.e.., ${\omega ^3} = 1,1 + \omega + {\omega ^2} = 0$, where $\omega ( \ne 1)$ is cube root of unity to solve the given problem.
We have given the equation, ${\left( {1 + {\omega ^4}} \right)^n} = {\left( {1 + {\omega ^8}} \right)^n}$.

Complete step-by-step answer:

We know that, ${\omega ^3} = 1,1 + \omega + {\omega ^2} = 0$. Using these formulas to the given equation,
\[
  {\left( {1 + {\omega ^4}} \right)^n} = {\left( {1 + {\omega ^8}} \right)^n} \\
   \Rightarrow {\left( {1 + \omega } \right)^n} = {\left( {1 + {\omega ^2}} \right)^n}\left[ {\because {\omega ^4} = {\omega ^3}.\omega = \omega ,{\omega ^8} = {\omega ^6}.{\omega ^2} = {\omega ^2}} \right] \\
   \Rightarrow {\left( { - {\omega ^2}} \right)^n} = {\left( { - \omega } \right)^n}\left[ {\because 1 + \omega = - {\omega ^2},1 + {\omega ^2} = \omega } \right] \\
   \Rightarrow {\omega ^{2n}} = {\omega ^n} \\
  for{\text{ n = 2,}} \\
   \Rightarrow {\omega ^{2 \times 2}} = {\omega ^4} = \omega \ne {\omega ^2} \\
  {\text{for n = 3}} \\
   \Rightarrow {\omega ^{2 \times 3}} = {\omega ^6} = 1 = {\omega ^3} \\
 \]
Hence, the value of least positive integer n is 3.

Note: After solving the equation, we have tried a hit and trial method in order to find the value of n. One might think, why we have not checked $n = 1$. This case is trivial. On $n = 1$, the equation will become ${\omega ^2} = \omega $, which is not true.