1. ## proof

w1 = $- 1/2 + √3/2 i$ and w2 =(w1)^2

show that [/FONT][/COLOR] $x^2 +xy+y^2=(x-w1y)(x-w2y)$

I am able to find (w1)^2
But that is all I can do

2. ## Re: proof

Ignore the orginal post
w1 = - 1/2 + (√3/2)i and w2 =(w1)^2

show that x^2 +xy+y^2=(x-w1y)(x-w2y)

I am able to find (w1)^2
But that is all I can do

3. ## Re: proof

So you know that $w_1= -1/2+ \sqrt{3}{2}i$ and $w_2= w_1^2= -1/2- \sqrt{3}/2 i$. To show the final result, work from right to left. $x- w_1y= x-(-1/2+ (\sqrt{3}/2)yi)$ and $x- w_2y= x-(-1/2-(\sqrt{3}/2)yi)$.

$(x- w_1y)(x- w_2y)= [(x+ (1/2)y)+ (\sqrt{3}/2}yi][(x+ (1/2)y)- (\sqrt{3}/2)yi$.

That is of the form $(a+ b)(a- b)= a^2- b^2$ with $a= x+ (1/2)y$ and $b= (\sqrt{3}/2)yi$ so is equal to $(x+ (1/2)y)^2- ((\sqrt{3}/2)yi)= x^2+ xy+ y^2/4- (3/4)(-y^2)= x^2+ xy+ y^2$.

4. ## Re: proof

I get it now thanks. I was stuck on
(a-b)^2 =(a-b)(a+b)=a^2-b^2

5. ## Re: proof

actually I am a bit confused.
x^2 +xy+y^2 is not in the form of a^2-b^2 so how can you let the components equal a^2-b^2

6. ## Re: proof

Originally Posted by markosheehan
I get it now thanks. I was stuck on
(a-b)^2 =(a-b)(a+b)=a^2-b^2
This is false. The second equality is true: $(a-b)(a+b) = a^2-b^2$, but the first is not. $(a-b)^2 = a^2-2ab+b^2 \neq a^2-b^2$.

7. ## Re: proof

Originally Posted by markosheehan
w1 = $- 1/2 + √3/2 i$ and w2 =(w1)^2
show that [/FONT][/COLOR] $x^2 +xy+y^2=(x-w1y)(x-w2y)$
I am able to find (w1)^2
But that is all I can do
Originally Posted by markosheehan
I get it now thanks. I was stuck on
(a-b)^2 =(a-b)(a+b)=a^2-b^2
Originally Posted by markosheehan
actually I am a bit confused.
x^2 +xy+y^2 is not in the form of a^2-b^2 so how can you let the components equal a^2-b^2
@markosheehan, if you are a bit confused, I am totally confused.

Please, please tell us all exactly what it is that you trying to do, show, or prove.

8. ## Re: proof

w1 = - 1/2 + (√3/2)i and w2 =(w1)^2

show that x^2 +xy+y^2=(x-w1y)(x-w2y)
$(x-w_1y)(x-w_2y) = x^2 - \color{red}{(w_1+w_2)}xy + \color{blue}{w_1w_2}y^2$

$\color{red}{w_1 + w_2} = w_1 + w_1^2 = \left(\dfrac{-1+i\sqrt{3}}{2}\right) + \left(\dfrac{-2-2i\sqrt{3}}{4}\right) = \dfrac{-2+2i\sqrt{3}-2-2i\sqrt{3}}{4} = \color{red}{-1}$

$\color{blue}{w_1w_2} = w_1 w_1^2 = \left(\dfrac{-1+i\sqrt{3}}{2}\right) \cdot \left(\dfrac{-2-2i\sqrt{3}}{4}\right) = \dfrac{2 +2i\sqrt{3}-2i\sqrt{3}+6}{8} = \color{blue}{1}$

9. ## Re: proof

I get it now.I am Sorry for being confusing