yet another complex cube roots (1 Viewer)

[cutemouse]

Hi,

Here's the question...

If p=a+b, q=aω+bω², r=aω²+bω

Prove that: p³+q³+r³=3(a+b)

Thanks,

Jason

 

[adnan91]

have u tried expanding?

 

[shanks27]

any more info?, because jst looking at it i would say expand and if you were given something like z^3 =1, u know that w^2 +w +1 =0 and so that lets u reduce all the higher powers of w to lower powers of w and then take them out as zero which should cancel alot of stuff out

 

[Js^-1]

Hey I tried expanding it, and i got:

p³+q³+r³ = 3a³+3a²b (1+w+w²) + 3ab²(1+w+w²) + 3b³

Now 1+w+w² = 0

So I got the answer to be:

P³+q³+r³= 3a³+3b³
=3(a³+b³)

Which is still not the answer...




[Edit]: Here's the working out to get p³+q³+r³ = 3a³+3a²b (1+w+w²) + 3ab²(1+w+w²) + 3b³

p=(a+b)
p³=a³+3a²b+3ab²+b³

q=w(a+bw)
q³=w³(a³+3a²bw+3ab²w²+b³)
q³=a³+3a²bw+3ab²w²+b³

Since w³=1​

r=w(aw+b)
r³=w³(a³w³+3a²w²b+3awb²+b³)
r³= a³w³+3a²w²b+3awb²+b³

Since w³=1​

p³+q³+r³ = a³+3a²b+3ab²+b³ + a³+3a²bw+3ab²w²+b³ + a³w³+3a²w²b+3awb²+b³
p³+q³+r³ = 3a³+3a²b (1+w+w²) + 3ab²(1+w+w²) + 3b³

 

[cutemouse]

Yes I did try expanding and got everything messed up.

What are some ways to efficently expand stuff like this?

Thanks,

Jason

 

[adnan91]

Yup thats what i got after expanding it and i have no clue as to where to go next. soz mate

 

[bored of sc]

Hi,

Here's the question...

If p=a+b, q=aω+bω², r=aω²+bω

Prove that: p³+q³+r³=3(a+b)

Thanks,

Jason

Hello Jason.

p³+q³+r³ = (p+q+r)³ -3(p²q+p²r+q²p+q²r+r²p+r²p+r²q)

Wait, that didn't help.

 

[shanks27]

Now 1+w+w² = 0


isn;t that only true if the question said z^3= 1 ??? thats why i was asking is there any parts before it like factors z^n= 1 ?

 

[joshlols]

Can you make ONE post and just repost your questions in there instead of making like 4 on questions you could've got simple advice from your teacher or something.

 

[Js^-1]

isn;t that only true if the question said z^3= 1 ??? thats why i was asking is there any parts before it like factors z^n= 1 ?

The complex 'n' th roots of unity are the solutions to zⁿ=1.

Aren't they?

 

[shanks27]

The complex 'n' th roots of unity are the solutions to zⁿ=1.

Aren't they?

yeh they are so for n=3 u were right in saying z^2 +z +1 will be 0, but it changes for each degrees of n basicall so that z^n -1 = (z-1) (z^n-1 +........+z+ 1)

 

[Js^-1]

Ok awesome thanks. I was seriously freaking right out.
So with the question...How do you get the correct answer? And where did I go wrong?

 

[Zeber]

omega is the complex root with the smallest argument.

 

[shanks27]

haha yeh. I am not sure about this question. I think it would of had an earlier part where u prove roots of unity otherwise it makes no sense.

So your mistake would of been assume w^3 equals 1, if u put that back in then not sure it might come out with some other parts. as i said it looks like a straight expand and algaebraic crunch when i worked it i got a fair way along but cbf typing it out, But i am thinking 1+w^3+w^6 will equal zero as that should get ride of a^3 and b^3 but my working. And then i think it might jst be some algreabic re arranging to take 3a +3b collectively out of the 3a^2b and 3ab^2.

And complex roots of unity is what all this omega stuff is. IT is the last or second last part of complex numbers at least it was for us