Can $\sqrt[4]{a} - \sqrt[4]{b}$ be rational? Let $a, b$ be two distinct positive integers such that both $\sqrt[4]{a}$ and $\sqrt[4]{b}$ are irrational. Can $\sqrt[4]{a} - \sqrt[4]{b}$ be a rational number? Should that be the case, I would be quite surprised!
NOTE. The analogous question with the fourth root replaced everywhere by the square root is easily seen to have a negative answer: see Iulu's answer to the post Integer Difference Between Roots. More generally Edge Erdil states in his answer to the post Is the difference of two irrationals which are each contained under a single square root irrational that the answer is negative if we replace the fourth root everywhere by the p-th root, with p prime.
 A: Describing the following case as an example for I think the question deserves an answer.
Assume that $a$ has a prime factor $p$ such that it appears with an odd multiplicity in the prime factorization of $a$. It follows that $x^4-a$ is irreducible over $\Bbb{Q}$. Either by Eisenstein or its stronger cousin — the Newton's polygon. Alternatively you can look at this older question and its answers.
In particular, this means that the field extension $K/\Bbb{Q}$, $K=\Bbb{Q}(\root4\of a)$ has degree four. A basis for $K$ as a vector space over $\Bbb{Q}$ consists of the elements $\mathcal{B}=\{1,a^{1/4},a^{1/2},a^{3/4}\}$.
Next we consider the number $s=\root4\of a-\root4\of b$. We assume contrariwise that $s$ is a rational number.
Should this happen it thus follows that $\root4\of b\in K$. Furthermore, raising to the fourth power gives the equation
$$
\begin{aligned}
b&=(\root4\of b)^4=(\root4\of a-s)^4\\
&=a-4a^{3/4}s+6a^{1/2}s^2-4a^{1/4}s^3+s^4\\
&=(a+s^4)-4s^3a^{1/4}+6s^2a^{1/2}-4sa^{3/4},
\end{aligned}
$$
where from last form we can read off the coordinates of $b$ with respect to the basis $\mathcal{B}$.
Because $b\in\Bbb{Q}$, uniqueness of the coordinates implies that all of $-4s$, $6s^2$, $-4s^3$ must vanish. In other words $s=0$, which is the uninteresting case $a=b$.

Galois theory comes to the fore, when we look at more complicated linear combinations of radicals with more than two terms, $\root4\of a+\root4\of b+\root4\of c$ and such. Mostly because it simplifies calculations of the degrees of the relevant field extensions. See the linked threads for more discussion.
Even in the case of a complicated linear combination of square roots Galois theory can be used. See this old answer of mine for an example. I want to emphasize that the use of Galois theory in that answer was a bit of a cutesy show-off. We could have equally well used the natural basis over $\Bbb{Q}$ of the relevant field extension, much the same way we did here.
