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.