Show $\mathbb{Q}[\sqrt[3]{2}]$ is a field by rationalizing I need to rationalize $\displaystyle\frac{1}{a+b\sqrt[3]2 + c(\sqrt[3]2)^2}$
I'm given what I need to rationalize it, namely $\displaystyle\frac{(a^2-2bc)+(-ab+2c^2)\sqrt[3]2+(b^2-ac)\sqrt[3]2^2}{(a^2-2bc)+(-ab+2c^2)\sqrt[3]2+(b^2-ac)\sqrt[3]2^2}$
But I fail to see the motivation behind this.  Can someone give a more general explanation of how to come to this expression?

You can multiply it by $1+d\sqrt[3]2+e\sqrt[3]4$ and insist the product not have any factors of $\sqrt[3]2$ or $\sqrt[3]4$.  So 
$$(a+b\sqrt[3]2+c\sqrt[3]4)(1+d\sqrt[3]2+e\sqrt[3]4)=\\a+ebe+2cd+\sqrt[3]2(ad+b+2ce)+\sqrt[3]4(ae+bd+c)$$
So we need $ad+b+2ce=0, ae+bd+c=0$, giving $$e=-\frac {ad+b}{2c}\\-a^2d-ab+2cbd+2c^2=0\\(a^2-2bc)d=2c^2-ab\\(a^2-2bc)e=b^2-ac$$  Then we make the leading term $a^2-bc$ to clear the fractions
