<s>[INST] Prove that $\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$ is a root for $x^3+\sqrt[3]{6}x^2-1$ 
Prove that 
  $$c=\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$$
  is a root for 
  $$P(x)=x^3+\sqrt[3]{6}x^2-1$$
  Source: list of problems for math contest preparation.

I have no clue on how to approach the problem. Most surely not by direct substitution, but I'm not seeing how. Some hint will be appreciated. 
 [/INST] I am going to expand the hint provided in the comments: by letting $\omega=e^{2\pi i/3}$, one if free to conjecture (especially if he/she knows how the Galois group of a cubic polynomial is usually structured) that the algebraic conjugates of 
$$ \alpha_1 = \frac{1}{\sqrt[3]{9}}\left[1+\sqrt[3]{-2}+\sqrt[3]{4}\right] $$
over $\mathbb{Q}(\sqrt[3]{6})$ are given by
$$ \alpha_2 = \frac{1}{\sqrt[3]{9}}\left[\omega+\sqrt[3]{-2}+\omega^2\sqrt[3]{4}\right] $$
$$ \alpha_3 = \frac{1}{\sqrt[3]{9}}\left[\omega^2+\sqrt[3]{-2}+\omega\sqrt[3]{4}\right]. $$
Let us see what the elementary symmetric functions of these $\alpha_1,\alpha_2,\alpha_3$ are.
$$ \alpha_1+\alpha_2+\alpha_3 = -\sqrt[3]{6} $$
$$ \alpha_1\alpha_2\alpha_3 = \frac{1}{9}\left[1^3+\sqrt[3]{-2}^3+\sqrt[3]{4}^3-3\sqrt[3]{1\cdot(-2)\cdot 4}\right]=1 $$
$$ \alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3 = \frac{3}{\sqrt[3]{9}^2}\left[\sqrt[3]{-2}^2-1\cdot\sqrt[3]{4}\right]=0$$
hence, yippie ya yeah, $\alpha_1,\alpha_2,\alpha_3$ are exactly the roots of $x^3+\sqrt[3]{6}x^2-1$.

As an alternative, it is possible to recall that $q(x)=x^3-6x^2+8$ is the minimal polynomial of $\sec\frac{2\pi}{9}$ over $\mathbb{Q}$. The algebraic conjugates of $\sec\frac{2\pi}{9}$ are $\sec\frac{4\pi}{9}$ and $\sec\frac{8\pi}{9}$, and the polynomials $x^3+\sqrt[3]{6}x^2-1$ and $q(x)$ are related via an affine map.
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