# Complex cubic roots of unity

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.

• 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. – 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 $$0. Then $$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$$.