I am given the fact that $\alpha^3=1$, where $\alpha \in \mathbb{C} \setminus \mathbb{R}$. With this in mind I have to find the value of the following equation:
$(1-\alpha)(1-\alpha^2)(1-\alpha^4)(1-\alpha^5)$
This is what I tried:
$\alpha^3=1$
$\alpha^3 - 1= 0$
$(\alpha - 1)(\alpha^2+\alpha+1)=0$; Solving this I found:
$\alpha \in \bigg\{1, -\dfrac{1}{2}+i\dfrac{\sqrt3}{2}, -\dfrac{1}{2}-i\dfrac{\sqrt3}{2} \bigg\}$.
However, I don't see how this might help. So I tried to simplify the equation as much as possible:
$(1-\alpha)(1-\alpha^2)(1-\alpha^4)(1-\alpha^5)=(1-\alpha)(1-\alpha^2)(1-\alpha^3 \alpha)(1-\alpha^3 \alpha^2)$
$=(1-\alpha)(1-\alpha^2)(1-\alpha)(1-\alpha^2)$
$=(1-\alpha)^2(1-\alpha^2)^2$
$=\bigg[ (1-\alpha)(1-\alpha^2) \bigg]^2$
$=(1-\alpha^2-\alpha+\alpha^3)^2$
$=(1-\alpha^2-\alpha+1)^2$
$=(2-\alpha^2-\alpha)^2$
$=...$
$=6- 3\alpha^2-3\alpha$
$=3(2-\alpha^2-\alpha)$
$=-3(\alpha^2+\alpha-2)$
$=-3(\alpha+2)(\alpha-1)$
Aaaand, I'm stuck. I think I'm missing some key insight since I keep getting stuck. Any ideas?
Since $\alpha^3=1$ we have $$(\alpha-1)(\alpha^2+\alpha+1)=\alpha^3-1=0,$$ but $\alpha- 1\ne0$ (because $\alpha\not\in\mathbb{R}$), so it must be that $\alpha^2+\alpha+1=0$.
Now, you have shown at one step that $$(1-\alpha)(1-\alpha^2)(1-\alpha^4)(1-\alpha^5)=(2-\alpha-\alpha^2)^2,$$ so writing $2-\alpha-\alpha^2=3-(1+\alpha+\alpha^2)=3$ we see $$(1-\alpha)(1-\alpha^2)(1-\alpha^4)(1-\alpha^5)=3^2.$$