Question

Someone explains me, please.

Answer 1

\(u^3+3u+3=(u+2)(u^2-2u+7)-11\)
hence \(1= \dfrac{1}{11}(u+2)(u^2-2u+7)-\dfrac{1}{11}(u^3+3u+3)\)
Clearly 1 = (the first term) + (- the second term)
But they let the first term =1 to get the inverse. I don't get

Answer 2

what are you saying? i'm sorry @Loser66

Answer 3

\[(u^3+3u+3)=(u+2)(u^2-2u+7)-11 \\ (u^3+3u+3)-(u+2)(u^2-2u+7)=-11 \\ \text{ divide both sides by } -11 \\ \frac{-1}{11}(u^3+3u+3)-\frac{-1}{11}(u+2)(u^2-2u+7)=1 \\ \frac{-1}{11}(u^3+3u+3)+\frac{1}{11}(u+2)(u^2-2u+7)=1 \\ \frac{1}{11}(u+2)(u^2-2u+7)-\frac{1}{11}(u^3+3u+3)=1 \]
are you asking how they came up with this?

Answer 4

No, no . Thanks for being here. I got it. :)

Answer 5

this sorta remembers me of finding the inverse like you know in number theory
like to find the inverse of a mod something...
\[\text{ Let } a^{-1} \text{ be the inverse of } a \text{ \in } \mod 7 \\ \text{ so if we wanted \to find } 2^{-1} \\ \text{ we would do } \\ 2a \equiv 1 (\mod 7) \\ 2a-1=7k \\ 2a-7k=1 \\ 7=2(3)+1 \\ 7-2(3)=1 \\ 2(-3)+7=1 \\ 2(-3)+7(1)=1 \\ 2(-3+7)+7(1-2)=1 \\ 2(4)+7(-1)=1 \\ \text{ so } 4 \text{ and } 2 \text{ are inverses of each other in } \mod 7\]

Answer 6

reminds me*

Answer 7

It is from ring theory, but apply number theory, it works well also. :)