Let $K$ be the splitting field over the rationals of some quartic with real roots and Galois group $S_4$, the symmetric group on 4 letters ($x^4-5x^3+6x^2-1$ will probably do). Now $S_4$ has a subgroup $G$ of order 8 (isomorphic to the group of symmetries of a square), and the fixed field $E$ of that subgroup will be a real cubic extension of the rationals. It is possible to compute a generator of $E$; whether it is trivial to do so is a somewhat subjective question. Personally, I'd say it's nontrivial, but I'm open to counterarguments.
EDIT: Let $a,b,c,d$ be the roots of the quartic. Then there are three cubic subfields, each generated by one of the numbers $ab+cd$, $ac+bd$, and $ad+bc$. These three numbers are conjugate over the rationals, and are the roots of the resolvent cubic, as Hurkyl asked, and as I forgot. There are formulas for the coefficients of the resolvent cubic in terms of the coefficients of the original quartic. The old theory-of-equations textbooks will discuss the resolvent cubic, as will the more computationally inclined of the abstract algebra texts, and of course it's bound to be discussed here and there on the web.