If $w^3=1$

Solve $(1-w)(1-w^2)(1-w^4)(1-w^5)$

I am not sure how to go about this question, I attempted expand the equation and simplify it using the basic roots of unity theorems, but it was . I am asking if anyone has a simple way of solving equations of these sort.

Thank you.

  • $\begingroup$ Use the following $1+\omega+\omega^2=0$ and $\omega^3=1$. For example using the second property, we can write $\omega^5=\omega^2$ and so on. $\endgroup$ – Anurag A Nov 12 '18 at 4:41

I presume $w\ne 1$. Then $1+w+w^2=0$. Also $w^4=w$ and $w^5=w^2$. Your product is $$[(1-w)(1-w^2)]^2=(2-w-w^2)^2=3^2=9.$$

More generally, let $\zeta$ be a primitive $n$-th root of unity. That is $\zeta^n=1$ and $\zeta^d\ne1$ for $0<d<n$. Then $$X^n-1=(X-1)(X-\zeta)(X-\zeta^2)\cdots(X-\zeta^{n-1}).$$ Dividing by $X-1$ and letting $X\to1$ gives $$n=(1-\zeta)(1-\zeta^2)\cdots(1-\zeta^{n-1}).$$ Your formula is the square of this with $n=3$.


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