My Math Forum I found new solution for this problem

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 May 5th, 2019, 11:20 AM #1 Newbie   Joined: Oct 2017 From: Egypt Posts: 15 Thanks: 0 I found new solution for this problem hi every on for omega "cubic roots for 1" what do you think ?
 May 5th, 2019, 01:31 PM #2 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 I think you need to clarify what you are doing. Thanks from topsquark
 May 6th, 2019, 12:13 PM #3 Newbie   Joined: Oct 2017 From: Egypt Posts: 15 Thanks: 0 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
 May 6th, 2019, 12:15 PM #4 Newbie   Joined: Oct 2017 From: Egypt Posts: 15 Thanks: 0 If you don't understand anything, see video or reply to me. Last edited by skipjack; May 6th, 2019 at 01:14 PM.
 May 6th, 2019, 12:42 PM #5 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 Small print is an obstacle.
 May 6th, 2019, 12:44 PM #6 Newbie   Joined: Oct 2017 From: Egypt Posts: 15 Thanks: 0 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.
 May 6th, 2019, 01:05 PM #7 Newbie   Joined: Oct 2017 From: Egypt Posts: 15 Thanks: 0 Please if you like my explanation hit like
 May 6th, 2019, 01:14 PM #8 Global Moderator   Joined: Dec 2006 Posts: 20,921 Thanks: 2203 As ω + ω² = -1, (1 - ω)(1 - ω²) = 1 - ω - ω² + ω³ = 2 + 1 = 3. Similarly, each successive pair of factors has product 3. Thanks from topsquark
 May 6th, 2019, 02:08 PM #9 Newbie   Joined: Oct 2017 From: Egypt Posts: 15 Thanks: 0 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.
 May 6th, 2019, 03:39 PM #10 Global Moderator   Joined: Dec 2006 Posts: 20,921 Thanks: 2203 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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