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