Question. Given roots $a,b,c$ of the cubic equation $x^3+3x^2+11x+1=0$, find $$s=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}.$$
Note that this polynomial is irreducible over $\Bbb Q$.
Vieta's formula was my first attempt while it goes extremely complicated after some computations. Also one could try to compute the minimal polynomial of $s$ but it might not be a easy job to determine which root is actually the value of $s$. I believe there should be a easier way out as this is from a high school homework question. The coefficients should be of significance, while I cannot observe anything special at this moment…
Attempt using Vieta's formula leads to $$(a^{1/3}+b^{1/3}+c^{1/3})^3-3(a^{1/3}+b^{1/3}+c^{1/3})((ab)^{1/3}+(bc)^{1/3}+(ac)^{1/3})-(a+b+c)+3(abc)^{1/3}=0, $$ or $$s^3-3((ab)^{1/3}+(bc)^{1/3}+(ac)^{1/3})s=0.$$ Then I have trouble computing $((ab)^{1/3}+(bc)^{1/3}+(ac)^{1/3})$…
off-topic: In the title "cube root of the roots of a cubic equation" looks weirdly symmetric as the pattern cube-root-roots-cubic…
Best Answer
A long comment. Regarding the cube root as a multi-valued function, $s$ can take $25$ different values, among which $0$ is the only "simple" one. Below depicts the location of all the possible values of $s$ on the complex plane:
All the other $24$ values seems rather complicated.
However, neither of the choice $\operatorname{Im}\log(z) \in (-\pi, \pi]$ nor $\operatorname{Im}\log(z) \in [0, 2\pi)$ for $\sqrt[3]{z} = e^{\frac{1}{3}\log z}$ yields $s = 0$. So I believe the question needs further clarifications.