If $\alpha=\sqrt[3]{2}$ and $p,q,r\in\mathbb{Q}$ then show $p+q\alpha+r\alpha^2$ is a subfield of $\mathbb{C}$ If $\alpha=\sqrt[3]{2}$ and $p,q,r\in\mathbb{Q}$ then show $p+q\alpha+r\alpha^2$ is a subfield of $\mathbb{C}$. 
For context, this is number $5$ in Chapter $1$ of Ian Stewart's Galois Theory. At this point in the text, we have only learned how to solve cubics and quartics, while introducing subring and subfield language. 
First to show $$R=\{p+q\alpha+r\alpha^2: p,q,r\in\mathbb{Q} \wedge \alpha=\sqrt[3]{2}\}$$ is a subfield  we show $R$ is a subring of $\mathbb{C}$ and then finish by showing $\forall x\in R,  \exists x^{-1}\in R$. 
Note that clearly $R\subset\mathbb{C}$ since $p+q\alpha+r\alpha^2$ is a real number for all rational $p,q,r.$ Take $p=1,q=0,r=0$ to see $1\in R$. 
If $p_1+q_1\alpha+r_1\alpha^2\in R$ and $p_2+q_2\alpha+r_2\alpha^2\in R$, then $$\left(p_1+q_1\alpha+r_1\alpha^2\right)+\left(p_2+q_2\alpha+r_2\alpha^2\right)=\left(p_1+p_2\right)+\left(q_1+q_2\right)\alpha+\left(r_1+r_2\right)\alpha^2\in R$$ $$-(p_1+q_1\alpha+r_1\alpha^2)=-p_2-q_2\alpha-r_2\alpha^2\in R$$ $$\left(p_1+q_1\alpha+r_1\alpha^2\right)\left(p_2+q_2\alpha+r_2\alpha^2\right)$$ $$=\left(p_1p_2+2q_1r_2+2q_2r_1\right)+\left(p_1q_2+p_2q_1+2r_1r_2\right)\alpha+\left(p_1r_2+q_1q_2+p_2r_1\right)\alpha^2\in R$$
The preceding argument follows from the facts that the rationals are closed under addition and multiplication. The above also shows $R$ is a subring of $\mathbb{C}$. To complete the proof that $R$ is a subfield, we find an expression for the inverse $$(p_1+q_1\alpha+r_1\alpha^2)^{-1}$$
Here is where I run into issues. My first thought was to set the product $\left(p_1+q_1\alpha+r_1\alpha^2\right)\left(p_2+q_2\alpha+r_2\alpha^2\right)$ equal to $1$: $$\left(p_1+q_1\alpha+r_1\alpha^2\right)\left(p_2+q_2\alpha+r_2\alpha^2\right)=1\implies$$ $$p_1p_2+2q_1r_2+2q_2r_1=1$$ $$p_1q_2+p_2q_1+2r_1r_2=0$$ $$p_1r_2+q_1q_2+p_2r_1=0$$ which is equivalent to $$\begin{pmatrix}
p_2&2r_2&2q_2\\ q_2&p_2&2r_2\\r_2&q_2&p_2\end{pmatrix}\begin{pmatrix}p_1\\q_1\\r_1\end{pmatrix}=\begin{pmatrix}1\\0\\0\end{pmatrix}$$
If I could find an explicit inverse for the above $3\times3$ matrix, the problem would be solved yielding exact expressions for $p_1,q_1, r_1$ in terms of $p_2,q_2,r_2$. 
However, I am not seeing a way to guarantee the determinant is nonzero for as long as $p_2,q_2,r_2\neq0$. 
I noticed the matrix is Toeplitz, but I don't know if that tells us anything about invertibility. 
Any help with finding the inverse element here without resorting to high power machinery in these answers Describe the subfields of $\mathbb{C}$ of the form: $\mathbb{Q}(\alpha)$ where $\alpha$ is the real cube root of $2$. and How to show that $\mathbb{Q}(\alpha) = \left\{ p+q\alpha+r\alpha^2 \mid p, q, r\in \mathbb{Q} \right\}$, where $\alpha$ is the real cube root of $2$? is much appreciated.
 A: Presumably, you know how to calculate the multiplicative inverse of complex numbers. This uses a similar idea to that, with rationalising the denominator. However, in this case there are three terms to the denominator, so it's a bit more tricky to find the exact term that works.
Note that
$$
(x+y+z)(x^2+y^2+z^2-xy-xz-yz)=x^3+y^3+z^3-3xyz
$$
Using this, we get
$$
(p+q\alpha+r\alpha^2)(p^2+q^2\alpha^2+2r^2\alpha-pq\alpha-pr\alpha^2-2qr)\\
=p^3+2q^3+4r^3-6pqr
$$
Now consider $\frac1{p+q\alpha+r\alpha^2}$ for some non-zero $p+q\alpha+r\alpha^2$, and expand this fraction according to the above. You now have a fraction with a rational number in the denominator, so it's in your ring.
Final piece: Showing that what we expand by is non-zero. We have that
$$
(x-y)^2+(x-z)^2+(y-z)^2\geq0\\
x^2+y^2+z^2- xy-xz-yz\geq0
$$
holds for any real $x,y,z$ with equality iff $x=y=z$. However, in our case that would mean
$$
p=q\alpha=r\alpha^2
$$
which by irrationality of $\alpha$ would imply $p=q=r=0$, which is not the case.
A: Hint: if $A$ is a finite-dimensional algebra over a field $K$ and $A$ is an integral domain, then $A$ is a field (because if multiplication by $x \neq 0$ is injective it must be surjective). (In your case, take $K = \Bbb{Q}$ and $A = \Bbb{Q}[\sqrt[3]{2}].)$
A: To help motivate the expressions for the conjugates, think about why conjugation rationalizes denominators over quadratic field extensions. Over $\mathbb{Q}[\sqrt{d}]$, $f(a+b\sqrt{d})\mapsto a-b\sqrt{d}$ is a field automorphism which fixes the rationals. For any $z\in\mathbb{Q}[\sqrt{d}]$, $zf(z)$ is fixed by $f$  (swapping the roots of $z^2-d$, $\sqrt d$ and $-\sqrt d$, just flips the order in the product), and so must be rational.
Instead of reaching for complex conjugation, you want an automorphism from a suitable permutation of the roots of the analogous irreducible (over $\mathbb{Q}$) polynomial, $z^3-2$. In this case, the automorphism from cyclically permuting the roots does what you want. Now you need three terms in the product to find a fixed element, which is why there are now two conjugates. 
So far, this guarantees that $\mathbb{Q}[\alpha, \omega\alpha, \omega^2\alpha]$ is a field, where $\omega$ is a third root of unity. If $g$ is the root permuting automorphism, you might be worried that the new "conjugate" of $z$, $g(z)g^2(z)$, lies outside of $\mathbb{Q}[\alpha]$. The same reasoning applied to complex conjugation shows that this "conjugate" must be real, which in this case is enough to conclude that it lies in the extension you care about.
A: This answer is for Carah's bounty, although I am also curious for a more detailed explanation of this answer. On the first linked question, a user named Daniel Juteau posted the following: (this is a direct copy-paste)
" The equation $X^3 - 2 = 0$ has three roots, namely $\alpha$, $j\alpha$ and $j^2\alpha$, where $j$ is a cubic root of unity. Therefore $p+q\alpha+r\alpha^2$ has two conjugates: $p+qj\alpha+rj^2\alpha^2$ and $p+qj^2\alpha+rj\alpha^2$. The product of the three will be rational. The inverse can be written:
$$\frac{1}{p+q\alpha+r\alpha^2}
= \frac{(p+qj\alpha+rj^2\alpha^2)(p+qj^2\alpha+rj\alpha^2)}
{(p+q\alpha+r\alpha^2)(p+qj\alpha+rj^2\alpha^2)(p+qj^2\alpha+rj\alpha^2)}
= \frac{(p^2-2qr)+(2r^2-pq)\alpha+(q^2-pr)\alpha^2}
{p^3+2q^3+4r^3-6pqr}$$
If you want to know more, you should read a course on Galois theory, for example Milne's lecture notes which are available online: http://www.jmilne.org/math/CourseNotes/FTe6.pdf "
It seems your professor (mine as well) both thought complex conjugation would work, but both of our professors couldn't determine the correct form. I  am interested in a detailed explanation of why those specific elements: $$p+qj\alpha+rj^2\alpha^2\text{ and }p+qj^2\alpha+rj\alpha^2$$ are conjugates of a given field element $p+q\alpha+r\alpha^2$. Thank you! 
