Someone plz solve this annoying q. (1 Viewer)

[Mathematician]

Given w, w^2 are the two complex roots of Unity.
Prove xyz = a^3 + b^3

if x = a + b
y= a +bw
z= a +bw^2

Im lost.

I got screwed in todays test ahhhhh


A Request:

Spice Girl and OldMan: Could u'se plz pick hard q's from cambridge or patel and solve them on the board everyday

Or A lot At the end of the week or when u have time because i think the examples from the books are always easy compared to the excercises and exams.


I need to study!!!
AHHH

 

[McLake]

Given w, w^2 are the two complex roots of Unity.
Prove xyz = a^3 + b^3

if x = a + b
y= a +bw
z= a +bw^2

(a + b)(a +bw)(a +bw^2)
= (a + b)(a^2 + abw + abw^2 + b^2w^3)
= (a^3 + a^2bw + a^2bw^2 + ab^2w^3 + a^2b + ab^2w + ab^2w^2 + b^3w^3)

don't know how to cancel ...

 

[spice girl]

Originally posted by Mathematician
Given w, w^2 are the two complex roots of Unity.
Prove xyz = a^3 + b^3

if x = a + b
y= a +bw
z= a +bw^2

z^3 - 1 = (z - 1)(z - w)(z - w^2)
(by definition of w, and the factor theorem)

then b^3 * (z^3 - 1) = (bz - b)(bz - bw)(bz - bw^2) .......(1)

Sub z = -a/b into (1)
b^3 * [ (-a/b)^3 - 1 ] = (-a - b)(-a - bw)(-a - bw^2) .......(2)

Multiply both sides of (2) by -1, and thus:

(a + b)(a + bw)(a + bw^2) = b^3 * [1 - (-a/b)^3]

= b^3 + a^3 as required

 

[enak]

Thanks spice girl.
Ey boctor, the test was relatively easy, but we need to study

 

[Mathematician]

thanks...

But why does z= -a/b ? howd u get it?
I cant think today.

And did u read the request ?
Hope im not asking too much

 

[Mathematician]

oh yeah..

oh yeah i was lost on that q, so i just did what Mc Lake did lol. HAHA

Do u think ill get any marks out of 4 for that q?

Kane- I got Hammered completly - I want my Revenge.

 

[spice girl]

Umm, you let z be -a/b, giving you the required result.

And umm...I'm really busy with uni stuff at the moment. May try once I settle down into the work.

 

[Mathematician]

...

I still dont get why u would let z= -a/b.
I know it gives the desired result .... so are u saying u can let z= (a^2+1)/(b+1) or something if u wanted to , but it wont get u anywhere?

My understanding is u can let z equal something as long as it doesnt have a and b from the z= a + bw^2


Also is there another way of doing this?

 

[Mathematician]

Ooops.

My understanding is u can let z equal something as long as it doesnt have a and b from the z= a + bw^2.


I meant since z is already expressed in terms of a and b.


btw sorry for posting too much.
Delete this after if u want.

 

[BlackJack]

Originally posted by McLake

= (a^3 + a^2bw + a^2bw^2 + ab^2w^3 + a^2b + ab^2w + ab^2w^2 + b^3w^3)

don't know how to cancel ...

[w, w^2 are 2 complex roots of 1...]
However, w.w^2 = w^3 = 1, and 1 + w + w^2 = 0

Hence,

= a^3 + a^2b ( w + w^2 + 1 ) + ab^2 ( w^3 + w + w^2 ) + b^3w^3.

= a^3 + a^2b ( 0 ) + ab^2 ( 0 ) + b^3 (1)

= a^3 + b^3.

The hard yakka way .

Oh yeah, I think there might be a need to say why I assumed 1 + w + w^2 = 0. Just add that in.

 

[McLake]

Thanks BJ, I knew I could do it my unnessiceraily long way ...

 

[spice girl]

Ahh...i get why you're lost Mathematician,

Umm, I shouldn't have chosen z. Let's choose t.

So since we're looking at a product xyz = (a+b)(a+bw)(a+bw^2), with a linear involving each of the cube roots of unity. The strategy was to recognise that it looks like a polynomial with roots 1, w, w^2.

So we start with the polynomial identity P(t) = t^3 - 1 = (t - 1)(t - w)(t - w^2)

And we screw around with P(t) until it gives the product. In this case, multiply by -b^3 and sub t = -a/b.

The trick is that we know that the product (t - 1)(t - w)(t - w^2)can be simplified to t^3 - 1 for all values of t.

 

[enak]

Re: oh yeah..

Originally posted by Mathematician
oh yeah i was lost on that q, so i just did what Mc Lake did lol. HAHA

Do u think ill get any marks out of 4 for that q?

Kane- I got Hammered completly - I want my Revenge.

Should get some marks for working

 

[Mathematician]

cool, thanx alll. I get it now.



thats a cool way spicegirl , but i think the other way will come to me quicker.

And Ill remember that. Maybe it will be usefull for another q.

 

[OLDMAN]

Students should spend some time analyzing Spice Girl's elegant solution , which he also used for the complex roots of unity question as well- it is a powerful approach.

A variant approach:

1,w,w^2 are roots of z^3 - 1=0

Find the equation whose roots are a+b,a+bw,a+b^2.

v=a+bz, therefore z=(v-a)/b.

Substituting, ((v-a)/b)^3 - 1 =0

1/b^3(v^3-3v^2a+3va^2-a^3)-1=0

from here product of roots, (a+b)(a+bw)(a+bw^2)=a^3+b^3