<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] So we have $\frac 1{1 + \sqrt[3]2}$ and we want to get 
$\frac 1{1 + \sqrt[3]2}\frac {something}{something}=\frac{something}{something\ with\ no\ radicals}$
Taking a page from doing this for square roots where, from $a + \sqrt b$ we realize $(a+\sqrt b)(a-\sqrt b) = a^2 - b$, which works because $(m-n)(m+n) = m^2 -n^2$.
If we use the idea $(m\pm k)(m^{n-1} \mp m^{n-2}k + ......) = m^n \pm k^n$. 
So if we can see $(1+\sqrt[3]{2})(1 - \sqrt[3]{2} + \sqrt[3]{2}^2) = 1 + \sqrt[3]2^3 = 1+2=3$
And $\frac 1{1 + \sqrt[3]2}=\frac 1{1 + \sqrt[3]2}\frac {1-\sqrt[3]2 + \sqrt[3]2^2} {1-\sqrt[3]2 + \sqrt[3]2^2}= \frac {1-\sqrt[3]2 + \sqrt[3]2^2}3=\frac 13 -\frac 13\sqrt[3]2 + \frac 13\sqrt[3]{2^2}$
Or $a = c =\frac 13$ and $b =-\frac 13$
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