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

Apply $a^3+1 = (a+1)(a^2-a+1)$, or 
$$\frac1{a+1}=\frac{a^2-a+1}{a^3+1}$$ to $a = \sqrt[3]{2}$ to obtain,
$$\frac1{1+\sqrt[3]{2}}= \frac{(\sqrt[3]{2})^2-\sqrt[3]{2}+1}{(\sqrt[3]{2})^3+1 }=\frac13(\sqrt[3]{2^2}-\sqrt[3]{2}+1) $$
