Polynomial/Complex number question! - need help! (1 Viewer)

[lisraeli]

Ok, I can't seem to be able to prove this one, a litle help would be good

parts 1 and 2 are ok, but you need them for 3) (nb, * is pie)

p(x) = x^6 + x^3 + 1

1. show roots of p(x)=0 are amongst x^9 - 1 = 0

2. Show p(x) = (x^2 -2xcos2*/9 + 1)(x^2 -2xcos4*/9 + 1)(x^2 -2xcos8*/9 + 1)

3. (cant get it)

hence show

cos2*/9.cos4*/9 + cos4*/9.cos8*/9 + cos8*/9cos2*/9 = -3/4


thanks a lot

 

[Riviet]

Basically we equate coefficients of x² using the result in 2:

1 + 1 + 1 + 4cos4*/9.cos2*/9 + 4cos2*/9.cos8*/9 + 4cos4/9.cos2*/9 = 0 (coefficient of x² in p(x) is zero)

Rearranging this gets you the required result.

 

[lisraeli]

ohhh thats how its done... thanks a heap. I was trying to do it using a AB + BC + AC style method (coefficients and relationships with roots).

 

[vafa]

This is a very popular questions.

solution:

View attachment 13976

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[lisraeli]

yeah i'm usually ok at them. I think the method of equatiing z^2 is far better then the method in the pictures

 

[Riviet]

yeah i'm usually ok at them. I think the method of equatiing z^2 is far better then the method in the pictures

I wasn't totally comfortable with it when I first read it in some book, but it's very useful and neat. vafa's method of substitution is also good if you can see what substitution you need.