<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] If you multiply both sides by $1+\sqrt[3]2$ you get $$1=(1+\sqrt[3]2)(a+b\sqrt[3]2+c\sqrt[3]4)\\
=a+2c+(a+b)\sqrt[3]2+(b+c)\sqrt[3]4$$
which we can resolve into three equations
$$1=a+2c\\0=a+b\\0=b+c$$
by  equating the parts proportional to $1,\sqrt[3]2,\sqrt[3]4$
which give 
$$b=-\frac 13\\
a=\frac 13\\c=\frac 13$$
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