Abstract Algebra (Multiplicative Inverse using the Euclidean Algorithm) The question I am trying to solve is to find the multiplicative inverse of 
$2-\sqrt[3]{2}+\sqrt[3]{4}$.
So far I am come up with the following.
Let $\alpha = 2-\sqrt[3]{2}+\sqrt[3]{4}$.
Then $s(x)=2-x+x^2$ and $x=\sqrt[3]{2}.$
Then $t(x)=x^3-2$.
The $gcd(s(x),t(x))=1$.
This implies there exists $\alpha(x),\beta(x)\in Q[x]$ such that $\alpha(x)s(x)+\beta(x)t(x)=1$.
Then $\frac{t(x)}{s(x)}=(x+1)+(\frac{-x-4}{2-x+x^2})$.
Then $(x^3-2)=(x+1)(2-x+x^2)+(-x-4)$ and $r_1=-x-4$.
Then $\frac{s(x)}{r_1(x)}=\frac{2-x+x^2}{-x-4}=(-x+5)+(\frac{22}{-x-4})$.
Then $2-x+x^2=(-x+5)(-x-4)+22$.
That is far as I have gone and I'm not sure if I am correct or where to go next.  
 A: The following Maple command 

gcdex(x^3-2,2-x+x^2,x,'s','t'); s; t;

will produce the gcd (1) and the factors $s(x),t(x)$ such that
$$
s(x)(x^3-2) +t(x) (2-x+x^2)=1. 
$$
You'll see that 
$$
s(x)=- 5/22 + 1/22 x
$$
and
$$
t(x)= 3/11 + 2/11 x - 1/22 x^2.
$$
Accordingly, 
$$
t(\sqrt[3]2) \cdot (2-\sqrt[3]2+\sqrt[3]4)=1,
$$
or
$$
(3/11 + 2/11 \sqrt[3]2 - 1/22 \sqrt[3]4) \cdot (2-\sqrt[3]2+\sqrt[3]4)=1.
$$
A: $\mathbb Q(\sqrt[3]{2} )$ is a field. Since $1, \sqrt[3]{2} , \sqrt[3]{4} $ form a basis for $\mathbb Q(\sqrt[3]{2} )$ over $\mathbb Q$, you know that
$$ (2-\sqrt[3]{2}+\sqrt[3]{4} )^{-1} = a + b\sqrt[3]{2}+ c\sqrt[3]{4} $$ 
for some $a, b, c \in \mathbb Q$.
One natural approach would be write the equation
$$ (2-\sqrt[3]{2}+\sqrt[3]{4} )( a + b\sqrt[3]{2}+ c\sqrt[3]{4}) = 1,$$
and expand the brackets.
If I've done this correctly, this gives
$$ (2a + 2b - 2c - 1) + (-a + 2b + 2c) \sqrt[3]{2}+ (a - b + 2c)\sqrt[3]{4} = 0.$$
But since $1, \sqrt[3]{2}$ and $ \sqrt[3]{4} $ are linearly independent over $\mathbb Q$, this means that the coefficients vanish individually:
$$ 2a + 2b - 2c = 1, \ \ \ -a + 2b + 2c = 0, \ \ \  a - b + 2c = 0.$$
Now you can go ahead and solve these simultaneous equations:
$$ a = \frac 3 {11}, \ \ \ b = \frac 2 {11}, \ \ \ c = - \frac 1 {22}, $$
So
$$ (2-\sqrt[3]{2}+\sqrt[3]{4} )^{-1} = \frac 3 {11} + \frac 2 {11}\sqrt[3]{2}+ - \frac 1 {22}\sqrt[3]{4} $$ 
