I read about casus irreducibilis here. As an example of casus irreducibilis, it says we can factor $x^3 - 15x - 4$ to find $4$ as a root and it also has two other real roots. Using Cardano's method we find $\sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i}$ as one of the roots. In fact $4 = \sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i}$ which can be confirmed by noting $(2 + i)^3 = 2 + 11i$ and $(2 - i)^3 = 2 - 11i$.
My question is, using Cardano's method for casus irreducibilis, is there any way to find out what the root actually is (without imaginary numbers) especially if it is a rational number. Is there a way other than guessing the explicit rational number beforehand (like in the example above). The only way I know to calculate the cube root is using Euler's equation but that will require you to know $\cos\left(\frac{\theta}{3}\right)$ and $\sin\left(\frac{\theta}{3}\right)$ but to write that in terms of $\cos(\theta)$ and $\sin(\theta)$ requires you to solve a cubic which starts an unending cycle (where $\theta = \tan^{-1}\left(\frac{b}{a}\right)$ for the complex number $a + bi$).