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.
On
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
$$c^3+\sqrt[3]6c^2-1=c^2(c+\sqrt[3]6)-1={\sqrt[3]4-1\over\sqrt[3]3}\cdot{1+2\sqrt[3]2+\sqrt[3]4\over\sqrt[3]9}-1={3\over3}-1=0$$
On
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.
On
Simplify $c$ before substituting to $P(x)=x^3+\sqrt[3]{6}x^2-1$: $$c=\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}=\frac{1-\sqrt[3]{2}+\sqrt[3]{2^2}}{\sqrt[3]{9}}=\frac{1+2}{\sqrt[3]{9}(1+\sqrt[3]{2})}=\frac{\sqrt[3]{3}}{1+\sqrt[3]{2}};\\ \begin{align}P\left(\frac{\sqrt[3]{3}}{1+\sqrt[3]{2}}\right)&=\frac{3}{(1+\sqrt[3]{2})^3}+\sqrt[3]{6}\cdot \frac{\sqrt[3]{9}}{(1+\sqrt[3]{2})^2}-1=\\ &=\frac{3+3\sqrt[3]{2}(1+\sqrt[3]{2})-(1+\sqrt[3]{2})^3}{(1+\sqrt[3]{2})^3}=\\ &=\frac{\color{red}3+\color{blue}{3\sqrt[3]{2}}+3\sqrt[3]{4}-\color{red}1-\color{blue}{3\sqrt[3]{2}}-3\sqrt[3]{4}-\color{red}2}{(1+\sqrt[3]{2})^3}=\\ &=0.\end{align}$$
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}