Proof about primitive roots of unity

Question

I need a proof for this:

Let $w$ be a primitive unity root order $3$. We need to prove that $(1-w +w^2)*(1+w-w^2) = 4$

Thanks.

Answer 1

Hint: If $w^3=1$ and $w\ne1$, then $\dfrac{w^3-1}{w-1}=w^2+w+1=0,$ so $w^2=-w-1$.

Let me know if you need further assistance.

Answer 2

We grind this one out as follows:

With

$w^3 = 1, \; w \ne 1, \tag 1$

we have

$w^4 = w^3 w = 1w = w, \tag 2$

and

$w - 1 \ne 0; \tag 3$

we observe that

$(w - 1)(w^2 + w + 1) = w^3 + w^2 + w - w^2 - w - 1 = w^3 - 1 = 0, \tag 3$

so in light of (3),

$w^2 + w + 1 = 0, \tag 4$

or

$-w^2 - w = 1; \tag 5$

then by virtue of (1), (2) and (5),

$(1 - w + w^2)(1 + w - w^2) = (1 + (w^2 - w))(1 - (w^2 - w)) = 1 - (w^2 - w)^2$ $= 1 - (w^4 - 2w^3 + w^2) = 1 - w^4 + 2w^3 - w^2$ $= 1 - w^4 + 2 - w^2 = 3 - w^4 - w^2 = 3 - w - w^2 = 3 + (-w - w^2) = 3 + 1 = 4. \tag 6$

Voila!