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May 7th, 2019, 07:21 AM   #21
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The given is three factors only
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May 7th, 2019, 07:24 AM   #22
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The proof you provided was for question (7) and it considered successive pairs.

Which question gives only three factors (no "..." to indicate further factors)?
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May 7th, 2019, 07:50 AM   #23
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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.
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July 8th, 2019, 07:49 AM   #24
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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
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
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 $ω^4=ω^3\times ω=1\times ω=ω$
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 $
Thanks $\color{red}{♥♥}$
Yahia kamal 2019

Last edited by yoyo145; July 8th, 2019 at 07:52 AM.
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(1-ω)(1-ω^2), cube roots of unity, cubic roots of unity, problem, yahia kamal

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