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
 A: Hint:
Rationalize the denominator by multiplying top and bottom by $1-\sqrt[3]2+\sqrt[3]4$
A: $\alpha=1+\sqrt[3]{2}$ 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.
A: 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 
$$b=-\frac 13\\
a=\frac 13\\c=\frac 13$$
A: 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 $b =-\frac 13$
A: 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,
$$\frac1{1+\sqrt[3]{2}}= \frac{(\sqrt[3]{2})^2-\sqrt[3]{2}+1}{(\sqrt[3]{2})^3+1 }=\frac13(\sqrt[3]{2^2}-\sqrt[3]{2}+1) $$
