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. 
 A: Let $\alpha = \sqrt[3]{\frac 19}$ and $\beta = 1-\sqrt[3]2+\sqrt[3]{2^2}$, so that $c = \alpha\beta$. Define $$Q(x) = P(\alpha x) = \frac{x^3}9+\frac{\sqrt[3]2}3 x^2-1$$ and note that $P(c) = 0$ iff $Q(\beta) = 0$.
Now, notice that $(1+\sqrt[3]2)(1-\sqrt[3]2+\sqrt[3]{2^2}) = 1^3+\sqrt[3]2^3 = 3$, so $\beta = \frac{3}{1+\sqrt[3]2}$.
Finally, substitute that into $Q(x)$:
\begin{align}
Q(\beta) &= \frac 19 \cdot\frac {27}{(1+{\sqrt[3]2})^3} + \frac{\sqrt[3]2}3\cdot\frac 9{(1+{\sqrt[3]2})^2}-1 \\
&=\frac 1{(1+{\sqrt[3]2})^2}\left(\frac{3}{1+\sqrt[3]2} +3\sqrt[3]2 - (1+{\sqrt[3]2})^2 \right)\\
&= \frac 1{(1+{\sqrt[3]2})^2}\left(1-\sqrt[3]2+\sqrt[3]{2^2} +3\sqrt[3]2 - 1-2{\sqrt[3]2} -\sqrt[3]{2^2}  \right) = 0.
\end{align}
A: It's not out of the question to compute
$$(1-\sqrt[3]2+\sqrt[3]4)^2=1+\sqrt[3]4+2\sqrt[3]2+2(\sqrt[3]4-\sqrt[3]2-2)=3(\sqrt[3]4-1)$$
so that
$$c^2=\left(1-\sqrt[3]2+\sqrt[3]4\over\sqrt[3]9 \right)^2={3(\sqrt[3]4-1)\over3\sqrt[3]3}={\sqrt[3]4-1\over\sqrt[3]3}$$
and, since $\sqrt[3]{54}=3\sqrt[3]2$,
$$c+\sqrt[3]6={1-\sqrt[3]2+\sqrt[3]4+3\sqrt[3]2\over\sqrt[3]9}={1+2\sqrt[3]2+\sqrt[3]4\over\sqrt[3]9}$$
Finally,
$$(\sqrt[3]4-1)(1+2\sqrt[3]2+\sqrt[3]4)=\sqrt[3]4+4+2\sqrt[3]2-1-2\sqrt[3]2-\sqrt[3]4=3$$
tells us that
$$c^3+\sqrt[3]6c^2-1=c^2(c+\sqrt[3]6)-1={\sqrt[3]4-1\over\sqrt[3]3}\cdot{1+2\sqrt[3]2+\sqrt[3]4\over\sqrt[3]9}-1={3\over3}-1=0$$
A: 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.
A: Simplify $c$ before substituting to $P(x)=x^3+\sqrt[3]{6}x^2-1$:
$$c=\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}=\frac{1-\sqrt[3]{2}+\sqrt[3]{2^2}}{\sqrt[3]{9}}=\frac{1+2}{\sqrt[3]{9}(1+\sqrt[3]{2})}=\frac{\sqrt[3]{3}}{1+\sqrt[3]{2}};\\
\begin{align}P\left(\frac{\sqrt[3]{3}}{1+\sqrt[3]{2}}\right)&=\frac{3}{(1+\sqrt[3]{2})^3}+\sqrt[3]{6}\cdot \frac{\sqrt[3]{9}}{(1+\sqrt[3]{2})^2}-1=\\
&=\frac{3+3\sqrt[3]{2}(1+\sqrt[3]{2})-(1+\sqrt[3]{2})^3}{(1+\sqrt[3]{2})^3}=\\
&=\frac{\color{red}3+\color{blue}{3\sqrt[3]{2}}+3\sqrt[3]{4}-\color{red}1-\color{blue}{3\sqrt[3]{2}}-3\sqrt[3]{4}-\color{red}2}{(1+\sqrt[3]{2})^3}=\\
&=0.\end{align}$$
