Rewrite $\frac{1}{1-\sqrt[3]{2}}$ as a polynomial question I've been looking for a way to rewrite the following fraction as a polynomial equation in $\sqrt[3]{2}$: $$\frac{1}{1-\sqrt[3]{2}}.$$ Now, to rewrite $1/(1-\sqrt{2})$ as a polynomial equation, it is enough to simply multiply by $(1+\sqrt{2})/(1+\sqrt{2})$, to get $$\frac{1}{1-\sqrt{2}} \frac{1+\sqrt{2}}{1+\sqrt{2}} = \frac{1+\sqrt{2}}{-1} = -(1+\sqrt{2}).$$ But this doesn't work with the other fraction (or maybe I'm missing something, haha). I've tried multiplying with different powers of two (or two raised to some fraction) to get the bloody $1/3$ power away, but it seems to persist. I've also tried successive multiplications of $(1+2^{1/3})/(1+2^{1/3})$ but that doesn't seem to bring anything either. My general ideas are to multiply by 1, choosing the correct fraction. Or doing the $+1$ and $-1$ trick, choosing $1$ appropriately. But I haven't been able to find the correct $1$. Any ideas?
 A: Write $\alpha = \sqrt[3]{2}$. We have $(\alpha-1)(\alpha^2+\alpha+1) = \alpha^3 - 1 = 1$, hence $\frac{1}{1-\alpha} = -\alpha^2 - \alpha - 1$.
A: Try multiplying by $$\frac{2^{2/3}+2^{1/3}+1}{2^{2/3}+2^{1/3}+1}$$
A: For a more general approach, let $\alpha$ be any algebraic number; suppose that $\alpha$ has minimal polynomial $p(x)$ and you want to evaluate $1/f(\alpha)$, where $f$ is a polynomial which is not a multiple of $p$.  (If $f$ is a multiple of $p$ then $f(\alpha)=0$ and so $1/f(\alpha)$ makes no sense.)
Use the Euclidean algorithm for polynomials to divide $p(x)$ by $f(x)$ and thereby to obtain an identity
$$1=p(x)a(x)+f(x)b(x)\ ,$$
where $a$ and $b$ are polynomials.  Substituting $x=\alpha$ gives by definition $p(\alpha)=0$ and therefore
$$\frac{1}{f(\alpha)}=b(\alpha)\ ,$$
which is what you want.
In your case you have $p(x)=x^3-2$ and $f(x)=1-x$.  Euclid works in one step,
$$x^3-2=(-x+1)(-x^2-x-1)-1$$
and so
$$1=(-x+1)(-x^2-x-1)-(x^3-2)\ .$$
Substituting $x=\alpha={\root3\of2}$ gives
$$1=(1-\alpha)(-\alpha^2-\alpha-1)-0=(1-\alpha)(-1-\alpha-\alpha^2)\ ,$$
and so
$$\frac{1}{1-\root3\of2}=-1-{\root3\of2}-({\root3\of2})^2\ .$$
