Example of $f \in K[x]$ solvable by radicals but having a root inexpressible only by coefficients of $f$ and +, -, *, /, $\sqrt[n]{...}$ Remark: at a very first glance it might seems as a duplicate of Example of $f \in K[x]$ solvable by radicals but having a root that cannot be expressed by using only coefficients of $f$, $+,-,\cdot,\frac{..}{..}$ but it is not. I made overlook in that question - I forgot to specify $\sqrt[n]{...}$ as an operation that we can use in an expression.
The definitions below are taken from Solvability by radicals implies a radical formula for its roots (question by Eparoh):

Definition 1: We say that a field extension $F/K$ is a radical extension if we can form a chain of fields
$$K=K_0 \leq K_1 \leq \cdots \leq K_n=F$$
where $K_{i+1}/K_i$ is a simple extension such that $K_{i+1}=K_i(a_i)$ and $a_i^{k_i} \in K_i$ for some positive integer $k_i$.
Definition 2: Let $K$ be a field and $f(x) \in K[x]$, we say that $f$ is solvable by radicals if there exists a radical extension $F/K$ such that $F$ contains a splitting field of $f$ over $K$.

This question has no answer, but it has a comment by reuns:

The radical formulas for the roots depend on constants of $K$, once the polynomial is fixed this is all we want (there are algorithms for the splitting field minimal polynomials and Galois group, if it is solvable we can unroll to find the radical formulas). What you are asking is if there are finitely many radical formulas $F_{d,l}$ of $d+1$ variables such that for every solvable polynomial $∑_{j=0}^{d} c_j x_j \in K[x]$ of degree $d$ its roots are given by $F_{d,l}(c_0,…,c_d)$ for some $l$. This is the problem of moduli space / parametrization of solvable polynomials of degree $d$.

Let $K$ be a field. Can you give an example of $f \in K[x]$ that is solvable by radicals but cannot be expressed by using only polynomial coefficients, $+, -, \cdot, \frac{...}{...}$ and operations of taking roots of natural ($N_+$) degrees and proof of this fact?
As I understand the comment I quoted above such polynomials and roots exist. I ask this question after a long quest for the answer to exactly the question asked in the linked post. I have no idea where to look for examples of such polynomials and roots in literature. To find the answer to the original question was itself hard. I have tried to google for "moduli space / parametrization of solvable polynomials" just having little hope that it will return information relevant to the posed problem, but no luck (as could be of course expected since it is not something directly related).
I must say that I do not exactly understand the quoted comment, but I think I will ask another question to resolve my doubts.
Edit 1: I have added this because it seems that the repliers make some implicit assumption which is exactly the point of my question. What I'm asking about is if there is an example of $f \in K[x]$ that is solvable by radicals but cannot be expressed by using ONLY polynomial coefficients, $+,−,\cdot,$ and operations of taking roots of natural ($N_+$) degrees. That is, it is not allowed to use these members of K that are not expressible in this form. The question from which I have taken definitions asks a very similar question, but it is not the same as it does not ask for such a specific example. I do not exactly understand the comment by reuns and I have some doubts if it is correct (please see the questions from which I took definitions, I think it will be beneficial for this discussion). But as I understand it, it says that if the assumptions I listed are satisfied then it is always possible to express roots using only members of K, $+, -, \cdot, \frac{...}{...}$ and operations of taking roots of natural ($N_+$) degrees, but not necessarily only polynomial coefficients, $+, -, \cdot, \frac{...}{...}$ and operations of taking roots of natural ($N_+$) degrees.
 A: I think this is more of a confusion of language and nothing else. If $f(x) \in K[x] $ is a specific polynomial then the coefficients of $f$ are nothing but specific members of $K$.
And then if you have a formula for roots of $f$ which involves a combination of some members of $K$ along with operations like $+, -, \times, /, \sqrt[n] {. } $ then the coefficients of $f$ themselves being members of $K$ can not be visually located in the formula. Any member of $K$ can for example easily be written as a combination of any given number of members of $K$ using just the field operations.
You are perhaps trying to think of an example where the coefficients are literals like in case of $x^2+ax+b$ and $K=\mathbb{Q} $, but again this is wrong. In such case the field should be $K=\mathbb{C} (a, b) $.
Let us then assume that we have a literal polynomial $$f(x)= x^n+a_1x^{n-1}+\dots+a_{n-1}x+a_n$$ over field $K=\mathbb{C} (a_1,a_2,\dots,a_n)$. If $f$ is solvable by radicals over $K$ then formula for roots involves arithmetic operations and radicals (nested if needed) applied on members of $K$ and it does include the literal coefficients of $f$ because they are what $K$ is made of. This is easily seen to be the case in case of quadratic or cubic equations which are known to be solvable.
Thus the coefficients always enter the formula for roots if there is a formula available.
Also note the well known fact (established by Abel well before Galois) that the polynomials with literal coefficients are solvable over their field of coefficients ($K=\mathbb{C} (a_1,a_2,\dots,a_n)$) if and only if $n<5$.

To summarize such an example which you are seeking does not exist.

I have tried to discern the meaning of the comment by reuns and it appears related to the treatment of solvable quintic given by Dummit and Foote in his Abstract Algebra.
They describe a criterion to check whether a given quintic $$f(x) =x^5+ax^4+bx^3+cx^2+dx+e\in\mathbb{Q}[x]$$ is solvable over $\mathbb{C} $. The idea is to form a complicated polynomial of degree 6 in $\mathbb{Q} [x] $ with coefficients made using coefficients of $f$ and checking whether it has a rational root or not.
And if the polynomial of degree 6 mentioned above does have a rational root then $f$ is solvable by radicals over $\mathbb{C} $. You perhaps want to check (for this case) if there is a formula for roots based on elements of $K=\mathbb {C} (a, b, c, d, e) $. I think there is such a formula but I am not sure.

Usually when we consider the problem of solvability of a polynomial $f(x) \in K[x] $, the field $K$ is the smallest field containing the coefficients of $f$. In this case if the polynomial is solvable by radicals over $K$ then the roots can be expressed in terms of coefficients of $f$ via arithmetic operations and radicals.
Enlarging the field $K$ to some extension $L$ and checking solvability over $L$ makes the problem simpler (trivial if $L$ is splitting field of $f$).
Also if we consider the scenario where $f(x) \in K[x] $ is solvable by radicals over $K$ and $F\subset K$ is the smallest field containing the coefficients we need to investigate the problem of solvability of $f$ over $F$ separately and one can't deduce anything from its solvability over $K$.
Thus your problem makes sense only in the usual setting where the solvability is checked over the field of coefficients and then (to repeat what I said earlier) the kind of example you seek does not exist.
