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 May 7th, 2019, 07:21 AM #21 Newbie   Joined: Oct 2017 From: Egypt Posts: 15 Thanks: 0 The given is three factors only
 May 7th, 2019, 07:24 AM #22 Global Moderator   Joined: Dec 2006 Posts: 21,015 Thanks: 2250 The proof you provided was for question (7) and it considered successive pairs. Which question gives only three factors (no "..." to indicate further factors)?
 May 7th, 2019, 07:50 AM #23 Newbie   Joined: Oct 2017 From: Egypt Posts: 15 Thanks: 0 I explain no.7 because my friend in school asked me to solve it but in my proof i answered that if we get only three factors like no.8 we able to solve it in convenient way and the student able to understand it Last edited by yoyo145; May 7th, 2019 at 08:04 AM.
 July 8th, 2019, 07:49 AM #24 Newbie   Joined: Oct 2017 From: Egypt Posts: 15 Thanks: 0 finally,i'm able to write on forum :) Hi, First of all, I would like to thank you for taking your time…… Prove that: $(1-ω)(1-ω^2)(1-ω^4)(1-ω^8$)……………………to 2n factors $=3^n$ In my problems book ,I found this problem. I tried to solve it by myself “as usual I can’t solve it at first time “ . so I saw my guide answer book . the answer wasn’t clear for me at first time “that is unusual ” I asked my teacher for that question but he didn’t add anything for me to understand it. so, I search on the internet for explanation but most of them are repeat what my book say !! I don’t lose my hope. I found someone asked the same question in this site https://math.stackexchange.com/quest...-of-2n-factors From the answers on this question I knew The 2n factors mean the” numbers of factors “ but I wrote it in my book as a note . I continue my study normally but my brain can’t stop thinking about that problem So, I decide to break up my problem ………….. I note $(1-ω)(1-ω^2)(1-ω^4)$ they are factors multiply to each other I began to search “how to multiply factors”…….after 4 hours I found that method to multiply factors $N^{no.of factors/2}$ Number of factors =2n “me: oh yeah!!!” What is n ?? we should study factors first "Factors" are the numbers we can multiply together to get another number: $2\times{3}$ $=6$ 2 and 3 are factors of 6 A number can have many factors. For example : $12 =$ $1\times{12}$ $=$ $2\times6$ $=$ $3\times4$ so all factors of “12” ={1,2,3,4,6,12} note it’s set.set should have all unique items. We can’t repeat numbers now I will tell you “n”=12 “the number that has factors” if we multiply these factors together we will find =1728 let’s use the formula $N^{no.of factors/2}$ we have 6 factors of 12 $12^{6/2}$ $=1728$ Then, the formula is valid Before we solve our problem remember these relation to help us simplify our problem $ω^3$=$1$ $1+ω+$$ω^2$$=0$ $-ω=1+$$ω^2 -ω^2=1+ω But , wait how to get “n” from our problem . we know we get it by multiply two factors But we can’t get 12 by multiply 1 and 2 “1&2 are factors of 12” we should find the right two factors to multiply to each other…..okay! Let’s solve our problem from this point Prove that: (1-ω)(1-ω^2)(1-ω^4)(1-ω^8)……………………to 2n factors =3^n (1-ω) will equal (1+1+$$ω^2)$$=(2+$$ω^2)$ $(1-ω^2)=(1+1+ω)=(2+ ω)$ The next bracket we will find $ω^4=ω^3\times ω=1\times ω=ω$ $(1-ω^4)=(1-ω)=(1+1+ω^2)=(2+ω^2)$ So let’s put them front of us $(2+ω^2) (2+ ω) (2+ω^2)$……etc. we will note that two bracket repeat…..”two factors repeat” So we have only two factors $(2+ω^2 )(2+ω)$ which they are equal to “n”…………..and number of factors is “2n” Let’s use multiply factors formula $N^{no.of factors/2}$ $((2+ω^2)(2+ω))^{2n/2}$ …………2 will go away with 2 $((2+ω^2)(2+ω))^n=(4+2ω+2ω^2+ω^3)^n=(4+2ω+2 ω^2+1)^n=(2+2+2ω+2ω^2+1)^n=(2+2(1+ω+ω^2)+1)^n$ $=3^n$ Thanks $\color{red}{♥♥}$ Yahia kamal 2019 Last edited by yoyo145; July 8th, 2019 at 07:52 AM.

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