Let $w$ be a primitive root of a unit of order 3, prove that $(1-w+w^2)(1+w-w^2)=4$

Question

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.

Answer 1

Note that $w^3=1$ and $1+w+w^2=0$ $$=(1-w+w^2)(1+w-w^2)$$ $1+w^2=-w$ and $1+w=-w^2$ $$=(-w-w)(-w^2-w^2)$$ $$=(2w)(2w^2)$$ $$=4w^3$$ $$=4(1)=4$$

Answer 2

We wanted to show that $$(1-w+w^2)(1+w-w^2)=4$$

After expansion, you show that it is equivalent to

$$1-w^2+1+1-w^4=4$$

$$3-w^2-w^4=4$$

which is equivalent to $$1+w^2+w^4=0$$

(remark: note that you made a typo and obtained $-1$ in the original post)

Since $w^3=1$,

this is equivalent to $$1+w^2+w=0$$

which is a true statement since

$$1+w+w^2=\frac{1-w^3}{1-w}=\frac{1-1}{1-w}=0$$

Answer 3

$\,4 - (1 - w + w^2) (1 + w - w^2) = (1 + w + w^2) (3 - 3 w + w^2)\,$ 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 $4.$