The answer is yes, curtesy of Ramanujan.
For nested cubic radicals of the form $\sqrt[3]{\sqrt[3]{A}-B}$, denote $C =\sqrt[3]{B^3-A}$ and construct the corresponding cubic polynomial below
$$R(x)=x^3 + \frac{B+2C}3x^2 - \frac{(B-C)(2B+C)}{27}x+ \frac{(B-C)^3}{729}$$
which, per Ramanujan, has the property that its roots $x_1$, $x_2$ and $x_3$ satisfy
$$\sqrt[3]{x_1}+ \sqrt[3]{x_2 }+ \sqrt[3]{x_3 } =\sqrt[3]{\sqrt[3]{A}-B} $$
Thus, denesting $\sqrt[3]{\sqrt[3]{A}-B}$ reduces to solving the cubic equation $R(x)=0$, a straightforward exercise. Take the example of
$\sqrt[3]{21\sqrt[3]{6}-17}$, with $B=17$ and $C= -37$. Its denesting polynomial is
$$x^3 -19x^2+6x + 216 =(x-18)(x-4)(x+3)$$
which results in
$$\sqrt[3]{21\sqrt[3]{6}-17}=\sqrt[3]{18}+\sqrt[3]{4}-\sqrt[3]{3}$$
Listed below are other examples admitting rational detestation, along with their respective polynomials:
$$
\begin{align}
\sqrt[3]{\sqrt[3]{2}-1}&=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49},
&\>\>\> &x^3 -\frac13 x^2-\frac2{27}x + \frac8{729}\\
\sqrt[3]{39\sqrt[3]{12}-19}&=\sqrt[3]{48}+\sqrt[3]{9}-\sqrt[3]{4},
&&x^3 -53x^2+204x + 1728\\
\sqrt[3]{7\sqrt[3]{20}-1} &=\sqrt[3]{\frac {100}9}+\sqrt[3]{\frac {16}9}-\sqrt[3]{\frac 59},
&& x^3 -\frac{37}3x^2+\frac{340}{27}x + \frac{8000}{729}\\
\sqrt[3]{93\sqrt[3]{30}+19}&=\sqrt[3]{180}+\sqrt[3]{25}-\sqrt[3]{6},
&&x^3 -199 x^2+3270 x + 27000 \\
\sqrt[3]{13\sqrt[3]{70}-17}&=\sqrt[3]{\frac{245}9}+\sqrt[3]{\frac{50}9} -\sqrt[3]{\frac{28}9},
&&x^3 -\frac{89}3 x^2+\frac{1330}{27} x + \frac{343000}{729}\\
\end{align}
$$