Showing the value the expression takes using another approach Let $1, \omega, \omega^{2}$ be the cube roots of unity. Then the product
$$
\left(1-\omega+\omega^{2}\right)\left(1-\omega^{2}+\omega^{2^{2}}\right)\left(1-\omega^{2^{2}}+\omega^{2^{3}}\right) \cdots\left(1-\omega^{2^{9}}+\omega^{2^{10}}\right)
$$ is equal to ?

what i considered was all 10 epxression can be written as $\frac{2}{1+w} * \frac{2}{1+w^2}... $ , From that expanding two term wise the below product i got all equal to 1 and hence product of the required sum being $2^{10}$ , is there a another way of doing it ? Like considering a polynomial which when given a value w will give that required multiplication , or might the product of $(w+1)(w^2+1)(w^4 +1) ...$ ?

 A: Note that, $\omega^2 + \omega +1=0$. Also note that,
\begin{equation*}
2^n \equiv \begin{cases}
            1 (\text{   mod } 3), \hspace{1cm} \text{if $n$ is even}\\
            2 (\text{ mod } 3), \hspace{1cm} \text{if $n$ is odd}
           \end{cases}
\end{equation*}
Establishing above statement is easy. Using this and the fact that $\omega^3 =1$ we get,
\begin{equation*}
\omega^{2^n} = \begin{cases}
               \omega, \hspace{1cm} \text{ if $n$ is even}\\
               \omega^2, \hspace{1cm}  \text{if $n$ is odd}\\
               \end{cases}
\end{equation*}
Let, $S_n = 1- \omega^{2^n}+ \omega^{2^{n+1}}$ for all $n \geq 0$. Then We required the find the value of $S$. Where,
\begin{equation*}
S = S_0.S_1.S_3.....S_9
\end{equation*}
But using the above discussion notice that,
\begin{equation*}
S_n = \begin{cases}
      1-\omega+\omega^2, \hspace{1cm} \text{if $n$ is even}\\
      1+\omega-\omega^2, \hspace{1cm} \text{if $n$ is odd}
      \end{cases}
\end{equation*}
So,
\begin{align*}
S&= S_0.S_1.S_2.....S_8.S_9\\
 &= (1-\omega+\omega^2)^5(1+\omega-\omega^2)^5\\
 &= (-2\omega)^5.(-2\omega^2)^5  &&[\omega^2 + \omega + 1 =0]\\
 &= (-2)^{10}. \omega^{15}\\
 &= 2^{10}. (\omega^3)^5\\
 &= 2^{10}
\end{align*}
So the answer is $2^{10}$.
