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. 

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}
