I found new solution for this problem


Forum Staff
Dec 2006
The proof you provided was for question (7) and it considered successive pairs.

Which question gives only three factors (no "..." to indicate further factors)?
Oct 2017
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:
Oct 2017
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 “ :D . 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/questions/2299230/what-is-meaning-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 https://www.youtube.com/watch?v=g4zTQveQuuY
$N^{no.of factors/2}$
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 :D
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: