Prove that the set $\mathbb{Q}[\alpha] = \{a+b\alpha+c\alpha^2 \mid a, b, c \in\mathbb{Q}\}$ is closed under addition and multiplication. 
Let $\alpha = 2^{1/3}$. 
  
  
*
  
*Prove that the set $\mathbb{Q}[\alpha] = \{a+b\alpha+c\alpha^2 \mid a, b, c \in\mathbb{Q}\}$ is closed under addition and multiplication. 
  
*Prove that if $z \in \mathbb{Q}[\alpha]$ is nonzero then there exists $z^{-1} $such that $z*z^{-1} = 1$

I have proved that $\mathbb{Q}[\alpha]$ is closed under addition, but I am having difficulty proving that it is closed under multiplication. 
With all coefficients in $\mathbb{Q}$, let $a_1 + b_1\alpha + c_1\alpha^2 $ and $a_2 + b_2\alpha + c_2\alpha^2 \in \mathbb{Q}[\alpha]$. When multiplied, these two elements yield a fourth degree polynomial, and I do not know how to prove that it is an element of $\mathbb{Q}[\alpha]$
 A: You should use the fact that $\alpha^3 = 2$. So then
$$\begin{align*}
(a_1+b_1\alpha + c_1\alpha^2)(a_2+b_2\alpha+c_2\alpha^2) &= (a_1a_2) + (a_1b_2+b_1a_2)\alpha
+(a_1c_2+b_1b_2+c_1a_2)\alpha^2\\
&\qquad\mathop{+} (b_1c_2+c_1b_2)\alpha^3 + (c_1c_2)\alpha^4\\
&= (a_1a_2) + (a_1b_2+b_1a_2)\alpha + (a_1c_2+b_1b_2+c_1a_2)\alpha^2\\
&\qquad\mathop{+} (b_1c_2+c_1b_2)2 + (c_1c_2)\alpha^3\alpha\\
&= \Bigl( (a_1a_2)+2(b_1c_2+c_1b_2)\Bigr) + (a_1b_2+b_1a_2)\alpha\\
 &\qquad\mathop{+} (a_1c_2+b_1b_2+c_1a_2)\alpha^2+ (c_1c_2)2\alpha\\
&\vdots
\end{align*}$$
etc.
A: Hint $\,\alpha^3 \in V\! = \Bbb Q\langle1,\alpha,\alpha^2\rangle\Rightarrow\,\color{#c00}\alpha V\subseteq V\,\Rightarrow\, \color{#0a0}{\alpha^2}V\subseteq \color{#c00}\alpha V\subseteq V\,$ so $\,(a_1\! +\! b_1\color{#c00}\alpha\! +\! b_2 \color{#0a0}{\alpha^2})V\subseteq V$
To show $\,f(\alpha)\in V\,$ is invertible, note $\,x^3-2\,$ is irreducible over $\,\Bbb Q\,$  so $\,\gcd(f(x),x^3-2)=1\,$ since $\,\deg f \le 2,\,$ so we can invert by using the gcd Bezout identity - just like we invert in $\,\Bbb Z_n,\,$ i.e
$$ \begin{align}  a(x) (x^3-2) +\ &b(x) f(x) = 1\\[.2em] \Rightarrow\ \bmod x^3\!-2\!:\ \ \ \ \ &b(x) f(x)\equiv 1\end{align}\qquad\ \ $$
which - when translated to our quotient ring - says $\, b(\alpha) f(\alpha) = 1\,$ in $\,\Bbb Q[\alpha]\cong \Bbb Q[x]/(x^3\!-2) $
