Finding the multiplicative inverse of an element in $\mathbb Q[x]/(x^3-2)$ I have a problem here that asks: "Express the multiplicative inverse of $1+2^{1/3}-3\cdot2^{2/3}$ as $a_0+a_1\cdot2^{1/3}+a_2\cdot2^{2/3}$."
I believe they are asking us to find it by utilizing the Euclidean algorithm. I am pretty confused about how to do this. I am also confused about the whole concept and how this relates to fields...
 A: Convince yourself that $\mathbb{Q}[x]/(x^3-2)$ is same as $\mathbb{Q}(\theta)$ where $\theta =2^{\frac{1}{3}}$.. 
Now, Inverse of $1+2^{\frac{1}{3}}-32^{\frac{2}{3}}$ is same as inverse of $1+\theta -3 \theta ^2$...
and then... 
as $\mathbb{Q}(\theta)$ is a field, inverse of $1+\theta -3 \theta ^2$ exists 
and as $\mathbb{Q}(\theta)$ is of degree $3$ every element is expressed as $a_0 +a_1\theta +a_2 \theta^2$.. 
In your language, every element can be written as  $a_0+a_1 2^{\frac{1}{3}} +a_2  2^{\frac{2}{3}}$.  
In particular, inverse of $1+2^{\frac{1}{3}} -3 2^{\frac{2}{3}}$ can also be written as $a_0+a_1 2^{\frac{1}{3}} +a_2  2^{\frac{2}{3}}$.. 
Can you proceed further?
Slight Extension: 
As $x^3-2$ is irreducible, g.c.d of this with any other polynomial will be $1$
in particular, g.c.d of $x^3-2, 1+2x-3x^2$ is $1$ 
i.e., I have two polynomials $m_x$ and $n_x$ in $\mathbb{Q}[x]$ such that :
$m_x.(x^3-2)+ n_x (1+2x-3x^2)=1$
As this is an identity, we would have $m_{\theta}.(\theta^3-2)+ n_{\theta} (1+2\theta-3\theta^2)=1$
But then, $\theta^3-2=0$
So, we would have $n_{\theta} (1+2\theta-3\theta^2)=1$
so, $n_{\theta}$ will then be inverse of $(1+2\theta-3\theta^2)$.

Now, What is $n_{\theta}$??

