(Major revision to incorporate new results in this MO cubic version.)
Note: All coefficients are in the rationals.
I. Cubic
In the linked post, it was shown that given a general cubic (via its depressed form),
$$x^3+px+q=0\tag1$$
and its roots $x_i$, then it is possible to find infinitely many rational $u$,
$$G_3 := (u \pm x_1)^{1/3}+ (u \pm x_2)^{1/3}+ (u \pm x_3)^{1/3} = {v}^{1/3}\tag2$$
such that $v$ is also a cubic root or even rational. Apparently, a necessary but not sufficient condition is that $u$ should be a rational point on,
$$u^3+pu+q=w^3\tag3$$
which, after a transformation, is an elliptic curve.
II. Quartic
By analogy, the quartic version should be,
$$G_4:=(u +x_1)^{1/4}+ (u +x_2)^{1/4}+ (u +x_3)^{1/4}+(u+x_4)^{1/4} = {v}^{1/4}\tag4$$
where the $x_i$ are the four solutions of a quartic and $v$ is at most a quartic root. By trial and error, I found the equation,
$$x^4+x^3+\big(\tfrac{n-3}{2}\big)^2x^2+x+1=0\tag5$$
and using $u=0$, then the $RHS$ of $(4)$ is $v=t^2$ which is a root of,
$$(t^2 - 8t + 6 - n)^2 = 4 n t^2\tag6$$
Thus $v$ is a quartic but, by judicious choice of $n$, can be a quadratic or even just a linear root.
Example:
Let $n=81$, then we have the irreducible,
$$x^4 + x^3 + 1521x^2 + x + 1=0$$
with complex roots $x_{1,2} \approx -0.49\pm38.99i$, and $x_{3,4} \approx -0.0003\pm0.025i$. Let $u=0$ in $(4)$ then,
$$x_1^{1/4}+x_2^{1/4}+x_3^{1/4}+x_4^{1/4} = (225)^{1/4}$$
or,
$$x_1^{1/4}+x_2^{1/4}+x_3^{1/4}+x_4^{1/4} = (413+52\sqrt{61})^{1/4}$$
depending on the $4$th roots used.
III. Quintic
Question: Is it possible to find a class of quintics that is irreducible (over $\mathbb{Q}$) yet have roots $x_i$ such that,
$$G_5:=x_1^{1/5}+x_2^{1/5}+\dots+x_5^{1/5} = v^{1/5}$$
and $v$ at most is a root of a quintic also with rational coefficients, or is there something in Galois theory that may be an obstruction?
