Show $\mathbb{Q}[\sqrt[3]{2}]$ is a field by rationalizing

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

You can also rationalize this by making use of the algebraic identity

$$x^3 + y^3 + z^3 - 3xyz \; = \; \left(x^2 + y^2 + z^2 - xy - xz - yz\right)\left(x+y+z\right)$$

You can view this as analogous to binomial square root rationalization, in which having a denominator of the form $x+y,$ where $x^2$ and $y^2$ are free of radicals, can be freed of radicals by multiplying both numerator and denominator by $x-y.$ In your situation, $x+y+z$ is what you have in the denominator and what you do is multiply both numerator and denominator by the "conjugate" $x^2 + y^2 + z^2 - xy - xz - yz.$ This identity doesn't take care of rationalizing a general trinomial made up of cube roots, by the way. For this there is a much more complicated identity that will work. See this 16 November 2010 AP-calculus post at Math Forum for a discussion of how to obtain such an identity.

Incidentally, this algebraic identity has a lot of uses, the best known probably being one of the ways of obtaining the cubic formula for solving cubic equations. This identity makes a lot of appearances in 19th century algebra texts and more about it can be found in the following sci.math thread I started on 23 April 2009: Factorization of a^3 + b^3 + c^3 - 3abc. I've assembled over a hundred old references to it since then, and one day I might write a historical survey of its mathematical and educational appearances in 19th century literature. It's also a standard identity for those serious about mathematical olympiad problems.

A few months after those sci.math posts I discovered the following short survey paper on this algebraic identity:

Desmond MacHale, My favourite polynomial, The Mathematical Gazette 75 #472 (June 1991), 157-165.

See also Mark B. Villarino's recent (2 January 2013) paper A cubic surface of revolution.

You rationalize by using Euclid's algorithm as used to prove Bézout's identity.

Let $\alpha = \sqrt[3]{2}$. We have obviously to assume $a+b\sqrt[3]2 + c(\sqrt[3]2)^2 = a+b\alpha + c \alpha^2 \ne 0$.

Consider the polynomial $0 \ne a + b x + c x^2 \in \mathbb{Q}[x]$. Since $x^3 - 2$ is irreducible in $\mathbb{Q}[x]$, for the gcd we have $(a + b x + c x^2, x^3 - 2) = 1$. Use Euclid's algorithm to find polynomials $u, v \in \mathbb{Q}[x]$ such that $(a + b x + c x^2) \cdot u + (x^3 - 2) \cdot v = 1$. Now evaluate this for $x = \alpha$ to obtain $(a + b \alpha+ c\alpha^2) \cdot u(\alpha) = 1$, so that \begin{equation} \frac{1}{a+b\sqrt[3]2 + c(\sqrt[3]2)^2} = \frac{1}{a + b \alpha+ c \alpha^2} = u(\alpha). \end{equation}