<s>[INST] Find $a,b$ and $c$ if $(1+\sqrt[3]{2})^{-1}$ in the form of $a+b\sqrt[3]{2}+c\sqrt[3]{2^2}$ Given that   

$$ (1+\sqrt[3]{2})^{-1} =a+b\sqrt[3]{2}+c\sqrt[3]{2^2}$$
  find the value of rationals $a,b,c.$

Solution I tried:  I tried to rationalize it, but I'm not getting the answer:
$$\frac{1}{(1+\sqrt[3]{2})}\times \frac{(1-\sqrt[3]{2})}{(1-\sqrt[3]{2})}$$
$$\frac{(1-\sqrt[3]{2})}{1-2^{\frac{2}{3}}}$$
 so doing so not getting answer; also, I tried to expand it, but we have condition that $(1+x)^n$ where $n$ is in fraction can be expandable only when $x < 1$, but here the cube root of $2$ is not less than $1.$ 
Thank you
 [/INST] $\alpha=1+\sqrt[3]{2}$ is a root of $(x-1)^3=2$, and so
$$
3 = \alpha^3 - 3 \alpha^2 + 3 \alpha = \alpha(\alpha^2 - 3 \alpha + 3)
$$
Therefore,
$$
\alpha^{-1}=\frac13(\alpha^2 - 3 \alpha + 3)
$$
Now use that
$$
\alpha^2 = 1+2\sqrt[3]2+\sqrt[3]4
$$
This solution works in general because every nonzero algebraic number is a root of a polynomial with nonzero independent term.
</s>