Find the inverse formula of $p+qa+ra^2$ Where $a= \sqrt[3]{2}$. I need to find the inverse formula for numbers of the form $p+qa+ra^2$. So simply, $\frac{1}{p+qa+ra^2}$ but I assume I need to rearrange it to look "nicer"(It technically isn't wrong as an "inverse"). I did something like $\frac{p-qa+ra^2}{(p+qa+ra^2)(p-qa+ra^2)}=\frac{p}{(p+qa+ra^2)(p-qa+ra^2)}-\frac{qa+ra^2}{(p+qa+ra^2)(p-qa+ra^2)}$ and thought I could simplify but not really, it looks overly and unnecesarily complicated. Thoughts? Or is $\frac{1}{p+qa+ra^2}$ all...? I cannot think of anything that would make it look better.

I am getting the same result as ThomasAndrews following a slightly different path.

Express the desired answer as the linear combination $u+va+wa^2$, and solve in rationals

$$(u+va+wa^2)(p+qa+ra^2)=1\\\ =(up+vra^3+wqa^3)+(uq+vp+wra^3)a+(ur+vq+wp)a^2.$$

You get the system

$$\begin{align}pu+2rv+2qw&=1,\\\qu+\ \ pv+2rw&=0,\\\ru+\ \ qv+\ \ pw&=0.\end{align}$$

The determinant is $p^3+2q^3+4r^3-6pqr$, which doesn't factor nicely.

The final solution is

$$\frac{p^2-2qr}\Delta-\frac{2r^2-pq}\Delta a+\frac{q^2-pr}\Delta a^2.$$