Rationalizing General Denominators I've read on several answers how to rationalize a fraction of the form
$$
\frac{1}{\sqrt{a_{1}} + \dots +\sqrt{a_{n}}}
$$
by multiplying top and bottom by the numbers
$$\varepsilon_{1}\sqrt{a_{1}} + \dots +\varepsilon_{n}\sqrt{a_{n}},$$
where $\varepsilon_{i} \in \left\{1,-1\right\}$, avoiding the current denominator, that is, avoiding the number that corresponds to $\varepsilon_{i} = 1$, for all $i$. In fact, it is possible to do it restricting ourselves to the family
$$\sqrt{a_{1}} + \varepsilon_{2}\sqrt{a_{2}} + \dots +\varepsilon_{n}\sqrt{a_{n}},$$
that is, ignoring the terms where $\varepsilon_{1} = -1$. My questions are the following ones:

*

*Is there a way of doing it involving less products (for n large)?

*Can somebody give me a book or reference that deals with the more general case
$$
\frac{1}{\sqrt[m]{a_{1}} + \dots +\sqrt[m]{a_{n}}}
$$
or, more generally,
$$
\frac{1}{\sqrt[m_{1}]{a_{1}} + \dots +\sqrt[m_{n}]{a_{n}}},
$$
including an explicit algorithm and (if possible) a proof of its correctness?

Thank you in advance!
 A: Let $\alpha$ be the denominator. What you desire is the minimal polynomial of $\alpha$ in $\mathbb Q[x]$, which is the lowest degree polynomial $P$ with rational coefficients where
$$P(\alpha)=0$$
which means
$$\frac1\alpha=\frac{P(0)-P(\alpha)}{P(0)\alpha}=Q(\alpha)$$
is a polynomial in $\alpha$ and involves the minimum amount of products. There are many methods for finding such polynomials. If you wish you can even ask WolframAlpha.
The particular case where only square roots appear in the denominator can be seen here, where you can see several other methods, and generalizes to higher order radicals via using $\varepsilon_i$ being roots of unity. With some more theory, you will find that this factorization is minimal unless the products of powers of the terms are related e.g. $\sqrt[3]3+\sqrt[3]9$ has degree $3$ instead of degree $9$ (compare this to this).
As an example, the minimal polynomial of $\alpha=\sqrt2+\sqrt[3]3$ is
$$P(x)=x^6-6x^4-6x^3+12x^2-36x+1$$
which gives us the rationalization
$$\frac1{\sqrt2+\sqrt[3]3}=\frac1\alpha=36-12\alpha+6\alpha^2+6\alpha^3-\alpha^5$$
leaving expansion of the RHS (if it's what you want) to the reader.
