If $\alpha=\sqrt[3]{2}$ and $p,q,r\in\mathbb{Q}$ then show $p+q\alpha+r\alpha^2$ is a subfield of $\mathbb{C}$

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.

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 $A = \Bbb{Q}[\sqrt[3]{2}].)$