w1 =$\displaystyle - 1/2 + √3/2 i $ and w2 =(w1)^2
show that [/FONT][/COLOR]$\displaystyle x^2 +xy+y^2=(x-w1y)(x-w2y) $
I am able to find (w1)^2
But that is all I can do
So you know that $\displaystyle w_1= -1/2+ \sqrt{3}{2}i$ and $\displaystyle w_2= w_1^2= -1/2- \sqrt{3}/2 i$. To show the final result, work from right to left. $\displaystyle x- w_1y= x-(-1/2+ (\sqrt{3}/2)yi)$ and $\displaystyle x- w_2y= x-(-1/2-(\sqrt{3}/2)yi)$.
$\displaystyle (x- w_1y)(x- w_2y)= [(x+ (1/2)y)+ (\sqrt{3}/2}yi][(x+ (1/2)y)- (\sqrt{3}/2)yi$.
That is of the form $\displaystyle (a+ b)(a- b)= a^2- b^2$ with $\displaystyle a= x+ (1/2)y$ and $\displaystyle b= (\sqrt{3}/2)yi$ so is equal to $\displaystyle (x+ (1/2)y)^2- ((\sqrt{3}/2)yi)= x^2+ xy+ y^2/4- (3/4)(-y^2)= x^2+ xy+ y^2$.
$(x-w_1y)(x-w_2y) = x^2 - \color{red}{(w_1+w_2)}xy + \color{blue}{w_1w_2}y^2$w1 = - 1/2 + (√3/2)i and w2 =(w1)^2
show that x^2 +xy+y^2=(x-w1y)(x-w2y)
$\color{red}{w_1 + w_2} = w_1 + w_1^2 = \left(\dfrac{-1+i\sqrt{3}}{2}\right) + \left(\dfrac{-2-2i\sqrt{3}}{4}\right) = \dfrac{-2+2i\sqrt{3}-2-2i\sqrt{3}}{4} = \color{red}{-1}$
$\color{blue}{w_1w_2} = w_1 w_1^2 = \left(\dfrac{-1+i\sqrt{3}}{2}\right) \cdot \left(\dfrac{-2-2i\sqrt{3}}{4}\right) = \dfrac{2 +2i\sqrt{3}-2i\sqrt{3}+6}{8} = \color{blue}{1}$