Rationalizing the denominator ${1\over n+\sqrt[k]{r}\pm\sqrt[2k]{r}} $ Let $k,n,r \in {\mathbb N}$, $k>1$, $r>1$, and assume that the denominators in all formulas are not zero. 
In the examples below, the numerators do not contain radicals, and the fractions are equal to 1:
$$
 \frac{r^2-(2n^2-4n+1)r+n^4}{(n+\sqrt{r}+\sqrt[4]{r})(n+\sqrt{r}-\sqrt[4]{r})
(n^2+(1-2n)\sqrt{r}+r)} =1;
$$
$$
 \frac{r^2+(2n-1)(n^2-4n+1)r+n^6}{\small (n+x^2+x)(n+x^2-x)
(n^2-n x+(1-n)x^2 -x^3+x^4)
(n^2+n x+(1-n)x^2 +x^3+x^4)} =1,
$$
$$
\mbox{ where } x=r^{1/6}.
$$
These examples suggest that the denominator of $\displaystyle{1\over n+\sqrt[k]{r}\pm\sqrt[2k]{r}} $ can be rationalized so that the new denominator is an integer of the form 
$$r^2+rP_k(n)+n^{2k}, \qquad\mbox{ where }  
P_k(n) \mbox{ is a polynomial in } n. \tag{1}
$$
Indeed this conjecture is not difficult to prove by induction when $k$ is a power of $2$.
Questions: Does $(1)$ hold for all $k$? Can we predict the coefficients of the polynomial $P_k(n)$?
For $k=2, \quad P_2(n)=-2n^2+4n-1$.
For $k=3, \quad P_3(n)=(2n-1)(n^2-4n+1)$.
 A: Yes to both questions, but I (personally) don't have a closed form for $P_k(n)$.
The following text assumes you are familiar with the terms of Galois theory.  If not, here's a very brief gloss:
A field is a set of numbers such that addition, subtraction, multiplication, and division are well-defined on that field.  Given a field $E\geq F$, and $\alpha\in E$, we write $F(\alpha)$ to mean the smallest subfield of $E$ containing $F$ and $\alpha$.  An automorphism is the function from a field to itself created by renaming all the elements.  A field is a splitting field of a polynomial if the polynomial factors (splits) into linear factors with coefficients in that field.  The conjugates of an element of a field are the values produced by evaluating all the automorphisms of that field on that element.  For each permutation of the roots of a polynomial $p(x)$ with coefficients in $F$, there is a corresponding automorphism of the splitting field of $p(x)$ (and vice-versa) preserving $F$, created by relabling the roots.  The irreducible polynomial of a number is the polynomial of smallest degree having that number as a root (which is unique).  
Now, onto the "proof" (with humorous headers!):
A Modest Proposal in the Swiftian Sense:
I'm going to assume that $\sqrt[2k]{r}$ cannot be simplified, just to make things simple.  My argument can be extended straightforwardly to the nonsimplified case.  
Let's start by extending $n$ to be rational, not just integer.  If we can show that the $P_k(x)$ appearing in (1) have integer coefficients, then we will end up  with a denominator in $\mathbb{Z}[r]$ for integer $n$ anyways.  
A Candidate for Rationalization:
Since $n+\sqrt[k]{r}\pm\sqrt[2k]{r}=n+(\sqrt[2k]{r})^2\pm\sqrt[2k]{r}$, we want to work in the a field extension $\mathbb{Q}(r)\leq\mathbb{Q}(\sqrt[2k]{r})$.  But the irreducible polynomial of $\sqrt[2k]{r}$ is $x^{2k}-1$, which does not split in $\mathbb{Q}(\sqrt[2k]{r})$.  So instead, we should deal with $n+\sqrt[k]{r}\pm\sqrt[2k]{r}$ in that splitting field, namely, $\mathbb{Q}(\zeta,\sqrt[2k]{r})$, where $\zeta$ is a $2k$th root of unity.  We know from elementary complex variables that the conjugates of $\sqrt[2k]{r}$ in that field are $\zeta^j\sqrt[2k]{r}$ for all $j\in\{0,1,\cdots,2k-1\}$.  Let the corresponding automorphisms of $\mathbb{Q}(\zeta,\sqrt[2k]{r})$ fixing $\mathbb{Q}(r)$ be $\{\sigma_j\}_j$.  Furthermore, for each $j$, let $F_j$ be the fixed field of $\sigma_j$.  Now, consider \begin{align*}
D&=\prod_{j=0}^{2k-1}{\sigma_j(n+\sqrt[k]{r}\pm\sqrt[2k]{r})} \\
&=\prod_{j=0}^{2k-1}{\left(n+\sigma_j(\sqrt[k]{r})\pm\sigma_j(\sqrt[2k]{r})\right)} \\
&=\prod_{j=0}^{2k-1}{(n+\zeta^{2j}\sqrt[k]{r}\pm\zeta^j\sqrt[2k]{r})}
\end{align*}  If we apply any automorphism $\sigma_l$ to $D$, $\sigma_l$ will end up permuting the factors of $D$, by Cayley's Theorem.  So $\sigma_l$ fixes $D$, which means $D$ is in the fixed field of $\sigma_l$.  Since $l$ was arbitrary, $D\in\bigcap_{l=0}^{2k-1}{F_l}$.  But since $\{\sigma_l\}_l$ constitute every automorphism of $\mathbb{Q}(\zeta,\sqrt[2k]{r})$ fixing $\mathbb{Q}(r)$, $\bigcap_l{F_l}=\mathbb{Q}(r)$.  
The Admirable Qualities of Our Candidate:
Hence $D\in\mathbb{Q}(r)$, and $D$ has $n+\sqrt[k]{r}\pm\sqrt[2k]{r}$ as a factor.  So if we multiply the numerator and denominator by all the other factors of $D$, we rationalize the denominator in $\mathbb{Q}(r)$.  But wait!  Suppose we now take $n$ an integer, as promised above.  Then we built $D$ out of elements of $\mathbb{Z}[\sqrt[2k]{r},\zeta]$, which is a ring.  So $D\in\mathbb{Z}[\sqrt[2k]{r},\zeta]\cap\mathbb{Q}(r)=\mathbb{Z}[r]$, as we have assumed that the radicals of $r$ cannot be simplified.  We now need to figure out the structure of $D$ as an element of $\mathbb{Z}[r]$.  
A Shortcut for Calculation:
Because we know $D\in\mathbb{Z}[r]$, we don't need to expand all the terms in our product for $D$.  Each root of $x^{2k}-r$ is linearly independent (over $\mathbb{Q}$, no less!) of the rest, so, when we expand and collect our product, the only monomials that do not cancel are the nonnegative integer exponents of $r$.  To rephrase, after multiplying this out, we get a polynomial in $r$, with coefficients polynomials in $n$.  
We have $2k$ factors in our definition of $D$; each has a term corresponding to $\left(r^{\frac{1}{2k}}\right)^0$ (namely, $n$) and term corresponding to $\left(r^{\frac{1}{2k}}\right)^2$, as well as a term with an intermediate exponent of $r^{\frac{1}{2k}}$.  Simply by degree considerations, $D$, as a polynomial in $\sqrt[2k]{r}$, this polynomial cannot have a power of $\sqrt[2k]{r}$ higher than $$\left(\left(r^{\frac{1}{2k}}\right)^2\right)^{2k}=r^{\frac{1}{2k}\cdot2\cdot2k}=r^2$$  Similarly, the polynomial cannot have a power of $r$ lower than $$r^{\frac{1}{2k}\cdot0\cdot2k}=1$$  Putting it all together, $D$ is of the form $Q_k(n)r^2+P_k(n)r+R_k(n)$ for some polynomials $Q_k(x),P_k(x),R_k(x)\in\mathbb{Z}[x]$.  
The Denoument, of Sorts:
Let's calculate out what those polynomials in $n$ are.  A term can only contribute to $Q_k(n)$, if, from each factor of $D$, we chose $\zeta^{2j}\sqrt[k]{r}$.  There is, of course, precisely one such term: $$\prod_j{\zeta^{2j}\sqrt[k]{r}}=r^2$$  So $Q_k(n)=1$.  Similarly, a term only contributes to $R_k(n)$ if we chose $n$ from each factor of $D$, so $r(n)=n^{2k}$.  And, finally, for the tricky part: $P_k(n)$.  There are many ways to contribute to $p_k(n)$: for each $l\in\{0,1,\cdots,k\}$, we choose terms of the form $\zeta^{2j}\sqrt[k]{r}$ from exactly $l$ of the factors, terms of the form $\pm\zeta^j\sqrt[2k]{r}$ from exactly $2(k-l)=2k-2l$ of the factors, and a term of the form $n$ from the remaining $l$ factors.  To make it easier to write this out formally, let $K=\{0,1,\cdots,2k-1\}$.  Then: \begin{align*}
P_k(n)&=\frac{1}{r}\sum_{l=0}^k{\prod_{\substack{J\subseteq I\subseteq K\\ \#(I)=2k-l\\ \#(J)=l}}{\left(n^l\prod_{j\in J}{\left(\zeta^{2j}\sqrt[k]{r}\right)}\prod_{j\in I\setminus J}{\left(\pm\zeta^j\sqrt[2k]{r}\right)}\right)}} \\
&= \frac{r}{r}\sum_{l=0}^k{\left(n^l\!\!\!\prod_{\substack{J\subseteq I\subseteq K\\ \#(I)=2k-l\\ \#(J)=l}}{\left(\prod_{j\in J}{\zeta^{2j}}\prod_{j\in I\setminus J}{(\pm\zeta^j)}\right)}\right)} \\
&= \sum_{l=0}^k{\left(n^l\!\!\prod_{\substack{J\subseteq I\subseteq K\\ \#(I)=2k-l\\ \#(J)=l}}{\left(\zeta^{2\operatorname{sum}{(J)}}(\pm1)^{\#(I\setminus J)}\zeta^{\operatorname{sum}{(I\setminus J)}}\right)}\right)} \\
&= \sum_{l=0}^k{\left(n^l\!\prod_{\substack{J\subseteq I\subseteq K\\ \#(I)=2k-l\\ \#(J)=l}}{\left(\zeta^{\operatorname{sum}{(J)}+\operatorname{sum}{(I)}}(\pm1)^{2(k-l)}\right)}\right)} \\
&= \sum_{l=0}^k{\left(n^l\!\prod_{\substack{J\subseteq I\subseteq K\\ \#(I)=2k-l\\ \#(J)=l}}{\zeta^{\operatorname{sum}{(J)}+\operatorname{sum}{(I)}}}\right)}
\end{align*}  Unfortunately, I can't quite close out this argument.  Determining these sums of roots of unity seems like a straightforward proposition, but I'm not good enough at combinatorics to do it myself.  So the best I can do is provide the above (non-closed) formula for $P_k(n)$.  If you (or anyone else on the web) can work it out, bully for you.  
