A way to get to the cube roots of the nested radicals in the question, thus getting to the three separate roots "buried" in the Cardano formula above, without using DeMoivre's formula to find the cube roots of the nested radicals is, first set
$x=-21$ and $y=-\frac{260}{9}i\sqrt{3}$ and (without going into how the equation below was derived) then find any real roots of the following equation
$$\frac{(-64a^9+(48x)a^6+((15(x)^2)-3(3y)^2)a^3+(x)^3)}{-64} = 0$$
This expands to
$$a^9 +\frac{63}{4}a^6 - \frac{74215}{64}a^3 +\frac{9261}{64} = 0$$
and as luck would have it has three rational roots $a_1=3; $ $a_2=\frac{1}{2}$; $a_3=-\frac{7}{2}$.
Next solve the equations
$$a_1^3+3a_1b_1^2 = -21$$
$$a_2^3+3a_2b_2^2 = -21$$
$$a_3^3+3a_3b_3^2 = -21$$
(where $-21$ is the value set as $x$) for $b_1$, $b_2$ and $b_3$, or $$b_1=\pm\frac{4i\sqrt{3}}{3}$$ $$b_2=\pm\frac{13i\sqrt{3}}{6}$$ $$b_3=\pm\frac{5i\sqrt{3}}{6}$$ So the three cube roots of $$\sqrt[3]{-\frac{260}{9}i\sqrt{3}-21}$$ are $a$+$b$ or
$$a_1+b_1=-\frac{4i\sqrt{3}}{3}+3$$
$$a_2+b_2=\frac{13i\sqrt{3}}{6}+\frac{1}{2}$$
$$a_3+b_3=-\frac{5i\sqrt{3}}{6}-\frac{7}{2}$$
The same method would be used to derive the cube roots of $$\sqrt[3]{\frac{260}{9}i\sqrt{3}-21}$$
Summarize all the calculations including the cube roots of of $$\sqrt[3]{\frac{260}{9}i\sqrt{3}-21}$$, (which we did not calculate here), to get the following roots of the equation
$$x_1=\left(-\frac{4i\sqrt{3}}{3}+3\right) +\left(\frac{4i\sqrt{3}}{3}+3\right) -3 = 3$$
$$x_2=\left(\frac{13i\sqrt{3}}{6}+\frac{1}{2}\right)+\left(-\frac{13i\sqrt{3}}{6}+\frac{1}{2}\right) -3 = -2$$
$$x_3=\left(-\frac{5i\sqrt{3}}{6}-\frac{7}{2}\right)+\left(\frac{5i\sqrt{3}}{6}-\frac{7}{2}\right)-3=-10$$ Multiplying the factors $$(x-3)(x+2)(x+10)=0$$ equals
$$x^3+9x^2-16x-60=0$$ the polynomial the question is seeking to derive.
Indeed, Cardano's formula for this cubic equation is the same as that presented in the original question and the three cube roots of the nested radicals in the equation are in the solutions found.
