<s>[INST] Find inverse of $[x+1]$ in factor ring $\mathbb{Q}[x]/\left\langle x^3-2 \right\rangle$ Find inverse of $[x+1]$ in factor ring $\mathbb{Q}[x]/\left\langle x^3-2 \right\rangle$. I remember that I need to use the extended Euclidean algorithm, but it has been some time, so I am a bit rusty. Thanks in advance!
Edit: I tried it and, by solving $(x+1)(ax^2+bx+c)=1$, with $x^3=2$, i got the solution for the inverse: $[x^2/3 -x/3 + 1/3]$, is the method and result correct?
 [/INST] Extended Euclidean algorithm is exactly right. Maybe 
$x^3 -2 = (x^2 -x +1) (x+1) - 3 $
$ x+1 \, \, = (-x/3)(-3) + 1$. This, we may write as $(-x/3)(-3) = (x+1) -1$
So that finally, $(x^3 - 2) (-x/3) = (x^2 -x +1) (x+1) (-x/3) - 3 (-x/3) $
The right side further becomes $(x^2 -x +1) (x+1) (-x/3) + (x+1) - 1 = (x+1) \big[p(x) \big] - 1 $. Thus $$ (x^3 -2) + 1 = (x+1)p(x) $$
</s>