Let $w$ be a primitive root of a unit of order 3, prove that $(1-w+w^2)(1+w-w^2)=4$ The title is the statement of the problem.
I did the following:
$(1-w+w^2)(1+w-w^2)=$
$1+w-w^2-w-w^2+w^3+w^2+w^3-w^4=$
$1-w^2+w^3+w^3-w^4=$
$1-w^2+1+1-w^4=4$, * then,by definition of primitive root of order 3*
$w^4+w^2-1=0$ and the problem is that from here I do not know how to continue. I tried to make the following change of variables: $z=w^2$ then $w^4+w^2-1=0$ becomes $z^2+z-1=0$ but it is not right because this equation has only two solutions and the other has four. So I do not know how to solve this problem.

$\,4 - (1 - w + w^2) (1 + w - w^2) = (1 + w + w^2) (3 - 3 w + w^2)\,$
where $\,1 + w + w^2\,$ is the irreducible polynomial for a primitive cube root of $1$.
Another way is to multiply the product by $\,1-w\,$ and expand
 $\,(1 - w + w^2) (1 + w - w^2)(1 - w) = 1 - w - w^2 + 3w^3 - 3w^4 + w^5.\,$ Now use the cube root property to get
 $\,w^3 = 1,\, w^4 = w,\, w^5 = w^2\,$ and
$\,1 - w - w^2 + 3 - 3w + w^2 = 4(1 - w).\,$ The original product must be $4.$
