Is there a quadratically closed field strictly between the quadratic closures of $\mathbb{Q}$ and $\mathbb{Q}(\sqrt[3]{2})$? Let $K$ be the quadratic closure of $\mathbb{Q}$, and $K'$ the quadratic closure of $\mathbb{Q}(\sqrt[3]{2})$. Is there a quadratically closed field $L$ strictly between $K$ and $K'$, i.e. such that $K \subsetneq L \subsetneq K'$?
It is a particular case of my previous question, so this is inspired by questions about ruler and compass constructions. If we could find some element $r \in K' \setminus K$ such that $\sqrt[3]{2}$ is of degree $3$ in $\mathbb{Q}(r)$, then the quadratic closure of $\mathbb{Q}(r)$ would not contain $\sqrt[3]{2}$, because the quadratic closure only adds elements whose degrees are powers of $2$, and therefore we could define $L$ as this quadratic closure. The degree of $r$ over $\mathbb{Q}$, if there is such a $r$, is a power of $2$, because we have
$$ [\mathbb{Q}(\sqrt[3]{2},r) : \mathbb{Q}] = [\mathbb{Q}(\sqrt[3]{2},r) : \mathbb{Q}(\sqrt[3]{2})] \cdot [\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}] = 2^n \cdot 3$$
so
$$ [\mathbb{Q}(r) : \mathbb{Q}] = \frac{[\mathbb{Q}(\sqrt[3]{2},r) : \mathbb{Q}]}{[\mathbb{Q}(\sqrt[3]{2},r) : \mathbb{Q}(r)]} = \frac{2^n \cdot 3}{3} = 2^n.$$
We also have the reverse implication: if there is a quadratically closed field $L$ strictly between $K$ and $K'$, then $\sqrt[3]{2}$ must be of degree $3$ over $L$, otherwise it would be of degree smaller than $3$, and there is no element of degree $2$ over $L$, so it would be contained in $L$, and for any element $r \in L \setminus K$, $\sqrt[3]{2}$ would be of degree $3$ over $\mathbb{Q}(r)$.
To sum up, the question reduces to: is there an $r$ which is of degree a power of $2$ over $\mathbb{Q}$ and which is not polyquadratic over $\mathbb{Q}$, but which is polyquadratic over $\mathbb{Q}(\sqrt[3]{2})$?
You can find the formulas for the roots of a quartic polynomial here. We see that the solutions are contained in the quadratic closure of $\mathbb{Q}(\sqrt[3]{2})$ if the intermediate parameter $f$ is an integer or if it is equal to $\sqrt[3]{2}$. So if we could find such an irreducible quartic over $\mathbb{Q}$ and such that its roots are not polyquadratic, we would conclude. But that becomes quite far fetched.
Edit: Now cross-posted (not by me) on Mathoverflow.
 A: The question has been answered positively on Mathoverflow in two different ways. I post here a copy of my solution:
Let $K$ be the quadratic closure of $\mathbb{Q}$, and $L$ be the quadratic closure of $\mathbb{Q}(\sqrt[3]{2})$. The roots of the polynomial $P(x) := x^4 + 36x + 54$ are not in $K$, but they are in $L$, and $P$ is irreducible because it is irreducible modulo $5$, hence any root $\alpha$ of it is of degree $4$, and the quadratic closure of $\mathbb{Q}(\alpha)$ is strictly between $K$ and $L$. (I used https://www.alpertron.com.ar/POLFACT.HTM for the calculations modulo 5.)
First, why are the roots of $P$ not in $K$? It is because mod $7$, it decomposes in $(x+6)(x^3+x^2+x+2)$, which are irreducible, and the discriminant of this polynomial is $-5038848$, which is not divisible by $7$, therefore by Dedekind's theorem there is a cycle of order $3$ in the Galois group of $P$ (Again, calculation modulo 7 with https://www.alpertron.com.ar/POLFACT.HTM, and calculation of the discriminant with https://planetcalc.com/8188/.)
Finally, it is in the quadratic closure of $\mathbb{Q}(\sqrt[3]{2})$, because we can look at the explicit formula for the roots of a quartic:

In our case, $a = b = 0$, therefore $e = \sqrt{(27c^2)^2-4(12d)^3} = 11664$, and $f = \sqrt[3]{27c^2 + e} = \sqrt[3]{2^6 \cdot 3^6} = 2^2 \cdot 3^2$. Except that, only square roots and $\sqrt[3]{2}$ appear in the formula, thus the roots are in $L$.
