Missing solutions from complex cubic root

From Cardano's work in the 1600s, we have this famous example of a cubic polynomial equation:

$$x^3-15x-4=0.$$

Plugging the coefficients into the Cardano-Tartaglia formula, we get an expression for the solutions that reduces to:

$$x=\sqrt[3]{2+\sqrt{-121}}+\sqrt[3]{2-\sqrt{-121}}\\=\sqrt[3]{2+i\sqrt{121}}+\sqrt[3]{2-i\sqrt{121}}.$$

We are here dealing with cubic roots of imaginary numbers, which are known to provide three solutions. In fact, all three solutions to this particular equation are real.

Now to my question. I could replace the imaginary numbers under the radicals with their polar forms and solve for all three solutions by equating arguments and modulii, respectively, from the subsequent cubic equations that can be extracted. But, when trying out a purely algebraic route, only one solution pops out with the other two solutions still "hidden":

$$x=\sqrt[3]{2+i\sqrt{121}}+\sqrt[3]{2-i\sqrt{121}}\\=\sqrt[3]{(2+i)^3}+\sqrt[3]{(2-i)^3}\\=2+i+2-i\\ =4$$

In this latter approach, where does the issue arrise that causes two solutions to not appear?

The number $$2+11i$$ has three cubic roots:

• $$\omega_1=2+i$$;
• $$\displaystyle\omega_2=(2+i)\left(-\frac12+\frac{\sqrt3}2i\right)=-1-\frac{\sqrt3}2+\left(-\frac12+\sqrt3\right)i$$;
• $$\displaystyle\omega_3=(2+i)\left(-\frac12-\frac{\sqrt3}2i\right)=-1+\frac{\sqrt3}2+\left(-\frac12-\sqrt3\right)i$$.

And the cubic roots of $$2-11i$$ are:

• $$\eta_1=2-i$$;
• $$\displaystyle\eta_2=-1+\frac{\sqrt3}2+\left(\frac12+\sqrt3\right)i$$;
• $$\displaystyle\eta_3=-1-\frac{\sqrt3}2+\left(\frac12-\sqrt3\right)i$$.

It turns out that $$\eta_1=\frac5{\omega_1}$$, that $$\eta_2=\frac5{\omega_3}$$, and that $$\eta_3=\frac5{\omega_2}$$. So, the roots of that cubic are:

• $$\omega_1+\eta_1=4$$;
• $$\omega_2+\eta_3=-2-\sqrt3$$;
• $$\omega_3+\eta_2=-2+\sqrt3$$.

$$x = \sqrt[3]{2+i\sqrt{121}}+\sqrt[3]{2-i\sqrt{121}}$$

This is misleading, because a non-zero complex number has $$3$$ cube roots, and the above makes it look like $$x$$ could take $$9$$ values between the different combinations of the two cube roots.

The better way to write the cubic formula is $$x = \sqrt[3]{2+i\sqrt{121}} + \dfrac{5}{\sqrt[3]{2+i\sqrt{121}}}$$ where the two cube roots are always chosen to be the same.
( The $$5$$ in the numerator comes from $$\left|2 + i \sqrt{121}\right|^2 = 2^2 + 121 = 125 = 5^3$$. )

$$\sqrt[3]{2+i\sqrt{121}}\;$$ [...] $$\; =\sqrt[3]{(2+i)^3}\;$$ [...] $$\; = 2+i\;$$ [...]

$$(2+i)$$ is indeed one of the cube roots of $$\sqrt[3]{2+i\sqrt{121}}$$, and the other two are $$\omega(2+i)$$ and $$\omega^2(2+i)$$ where $$\omega = \dfrac{-1 \pm i \sqrt{3}}{2}$$ is a primitive cube root of unity. Then the three solutions are:

$$x = \omega^k(2+i) + \frac{5}{\omega^k (2+i)} = \omega^k(2+i) + \overline{\omega}^k (2 - i) = 2 \text{Re}\left(\omega^k (2+i)\right) \;\;\Big|\;\; k = 0, 1, 2$$