Showing the value the expression takes using another approach

Question

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 $(w+1)(w^2+1)(w^4 +1) ...$ ?

Answer 1

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 $2^{10}$.