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. 

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
$$c^3+\sqrt[3]6c^2-1=c^2(c+\sqrt[3]6)-1={\sqrt[3]4-1\over\sqrt[3]3}\cdot{1+2\sqrt[3]2+\sqrt[3]4\over\sqrt[3]9}-1={3\over3}-1=0$$
