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May 5th, 2019, 11:20 AM   #1
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I found new solution for this problem

hi every on
for omega "cubic roots for 1"

what do you think ?
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May 5th, 2019, 01:31 PM   #2
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I think you need to clarify what you are doing.
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May 6th, 2019, 12:13 PM   #3
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okay

the links
https://math.stackexchange.com/quest...-of-2n-factors


1.jpg

2.jpg

3.jpg

4.jpg

5.jpg

6.jpg

Yahia Kamal 2019
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May 6th, 2019, 12:15 PM   #4
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If you don't understand anything, see video or reply to me.

Last edited by skipjack; May 6th, 2019 at 01:14 PM.
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May 6th, 2019, 12:42 PM   #5
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Small print is an obstacle.
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May 6th, 2019, 12:44 PM   #6
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Hmm, I can't type omega or power in your forum??

You can save them and zoom.

Or open in new tape and zoom.

What is your opinion?

Last edited by skipjack; May 6th, 2019 at 01:15 PM.
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May 6th, 2019, 01:05 PM   #7
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Please if you like my explanation hit like
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May 6th, 2019, 01:14 PM   #8
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As ω + ω² = -1, (1 - ω)(1 - ω²) = 1 - ω - ω² + ω³ = 2 + 1 = 3.
Similarly, each successive pair of factors has product 3.
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May 6th, 2019, 02:08 PM   #9
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But why you are taking successive pairs?

I think 2n means the number of our factors.

Last edited by skipjack; May 6th, 2019 at 03:38 PM.
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May 6th, 2019, 03:39 PM   #10
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As there are 2n factors, there are n pairs of factors.

I used successive pairs of factors because each such pair has product 3.
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(1-ω)(1-ω^2), cube roots of unity, cubic roots of unity, problem, yahia kamal



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