cube roots of unity question (1 Viewer)

[Lurch]

Given 1, w & w^2 are the cube roots of unity.
How do we show that if z^3-1=0 & pz^5+qz+r=0 have a common root then (p+q+r)(pw^5+qw+r)(pw^10+qw^2+r)=0

 

[turtle_2468]

Sub in the 3 possible values of z.
basically, if z= one of 1,w,w^2
and we call the second poly P(z)
then one of P(1), P(w), P(w^2) is equal to zero
ie P(1)P(w)P(w^2)=0. Sub in P for answer.
This is funny... first answer for ages, and after I did a polynomials lecture this arvo as well..

 

[Lurch]

God how stupid was I doh...
thanks..

 

[KeypadSDM]

It's a silly question ... they should have simplified at the end to make it a tad more confusing.

 

[turtle_2468]

Complexity a good question doth not necessarily make...

 

[KeypadSDM]

Originally posted by turtle_2468
Complexity a good question doth not necessarily make...

But a harder one. This one was too leading.

I mean, come on, w¹⁰?

 

[CM_Tutor]

Originally posted by KeypadSDM
But a harder one. This one was too leading.

I mean, come on, w¹⁰?

So, would you have considered it a better question if it asked you to prove that:

(p + q + r)(pw² + qw + r)(pw + qw² + r) = 0

 

[KeypadSDM]

Originally posted by CM_Tutor
So, would you have considered it a better question if it asked you to prove that:

(p + q + r)(pw² + qw + r)(pw + qw² + r) = 0

I never sait better. I said the other question was silly, and I said this one would have been a bit more confusing.

Don't put words in my mouth.

 

[CM_Tutor]

Sorry, I wasn't trying to put words into your mout, I was simply asking a question.

 

[KeypadSDM]

Originally posted by CM_Tutor
Sorry, I wasn't trying to put words into your mout, I was simply asking a question.

It does look a little neater though.

 

[MyLuv]

This 1 is weird...Cant we just say if w is common root of z^3=1 and pz^5+qz+r then pw^5+qw+r=0
so we can just multiply any thing to it and the answer still 0
ie (p+q+r)(pw^5+qw+r)(pw^10+qw+r)=0
...???

 

[CM_Tutor]

Originally posted by MyLuv
This 1 is weird...Cant we just say if w is common root of z^3=1 and pz^5+qz+r then pw^5+qw+r=0
so we can just multiply any thing to it and the answer still 0
ie (p+q+r)(pw^5+qw+r)(pw^10+qw+r)=0
...???

No, we can't, as the question states that there is a common root, not that the common root is z = w. It might be
z = 1, or z = w², if we take w = cos(2pi/3) + isin(2pi/3)

Originally posted by KeypadSDM
It does look a little neater though.

I agree - in fact, I think it would be a better question

 

[MyLuv]

lol..i misread the 1st line (1,w,w^2 is root of unity )...thought w is a general number ^^