Find the inverse formula of $p+qa+ra^2$

Question

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.

Answer 1

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

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

Answer 2

This is the sneaky "linear algebra" approach.

Consider that $p+qa+ra^2$ acts on the field of values $u+va+wa^2$ by multiplication. This is a linear transformation, so can be written as matrix when written as an action on the normal basis $1,a,a^2$:

$$A=\begin{pmatrix}p&q&r\\2r&p&q\\2q&2r&p\end{pmatrix}$$

Find the characteristic polynomial of this matrix:

$$p(\lambda)=\det(\lambda I-A)=\lambda^3 +c_2\lambda^2+c_1\lambda+c_0$$

You will get that $c_0=-\det A\neq 0$, will be rational (assuming the unstated, that $p,q,r$ are rational,) and we have that $p(A)=0$, so you have that:

$$A(A^2+c_2A+c_1I)=-c_0I$$

Divide both sides by $-c_0$ and you get your result, so the inverse will be:

$$A^{-1}=\frac{-1}{c_0}A^2 + \frac{-c_2}{c_0}A + \frac{-c_1}{c_0}I.$$

Compute this out, and the top row will be the coefficients for the inverse.

The characteristic polynomial can be computed by hand: $$\begin{align} p(\lambda)&=(\lambda-p)^3 +2(-q)^3 + 4(-r)^3-6(\lambda -p)(-p)(-q)\\ &= \lambda^3 - 3p\lambda^2 + (3p^2-6qr)\lambda -(p^3+2q^3+4r^3-6pqr) \end{align} $$ That gives you $-c_0=p^3+2q^3+4r^3-6pqr, c_1=3p^2-6qr, c_2=-3p$.

The rest is just a grind.

Answer 3

$\newcommand{\Q}{\mathbb{Q}}$The standard approach to such a rationalization is to use Euclid's algorithm.

Consider the polinomial $f(x) = p + q x + r x^{2} \in \Q[x]$. You are assuming $f(a) \ne 0$, where $a = \sqrt[3]{2}$. Therefore $\gcd(f(x), x^{3} - 2) = 1$. Use Euclid's algorithm to find polynomials $g(x), h(x)$ such that $$ f(x) \cdot g(x) + h(x) \cdot (x^{3} - 2) = 1. $$ Substitute $x = a$ to get that $f(a) g(a) = 1$, that is $$ g(a) = \frac{1}{f(a)} = \frac{1}{p + q a + r a^{2} } $$