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OpenStudy (anonymous):
Perform the Indicated operation
(x+1)^3 in Z3[x]

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OpenStudy (anonymous):
\[ (x+1)^{3} in Z_{3}[x]\]

terenzreignz (terenzreignz):
\[\huge (x+1)^3 \ \ \ \ in \ \mathbb{Z}_3[x]\]
LOL any idea how to do this? :)

OpenStudy (anonymous):
do i just use f(0) and f(1) to solve for them in Z3? I'm not sure really thats why I'm asking and my book doesn't even address it.

OpenStudy (anonymous):
hint \(3\equiv 0\)

terenzreignz (terenzreignz):
Freshman's dream :D
Anyway, evaluate this
(x+1)^3 normally first, and tell me what you get.

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OpenStudy (anonymous):
lol

terenzreignz (terenzreignz):
Cube of a binomial?

OpenStudy (anonymous):
1+3 x+3 x^2+x^3

terenzreignz (terenzreignz):
That's right :)
\[\huge x^3 + 3x^2+3x+1\]but... we take everything in mod 3...
In particular, in
\[\large \mathbb{Z}_3\]
3=0, right?

OpenStudy (anonymous):
yes

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OpenStudy (anonymous):
so is it X^3 + 1?

terenzreignz (terenzreignz):
You got it :)
And now, for any prime number p

terenzreignz (terenzreignz):
\[\huge (a+b)^p=a^p+b^p \ \ \ \ \ in \ \mathbb{Z}_p\]

OpenStudy (anonymous):
i always suspected my math teachers were lying, and \((a+b)^3=a^3+b^3\)

terenzreignz (terenzreignz):
Oh abstract algebra... so deceptive :D
That was fun... but I'm getting drowsy :/
-----------------------------------------
Terence out

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OpenStudy (anonymous):
Thank you

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