TITLE: Rational solutions to x^3 + y^3 + z^3 - 3xyz = 1
QUESTION [5 upvotes]: I can show that there infinitely many solutions to this equation. Is it possible that the set
of rational solutions is dense?

REPLY [6 votes]: The answer is yes, the rational points on your surface lie dense in the real topology.
Let's consider the projective surface $S$ over $\mathbb{Q}$ given by $X^3+Y^3+Z^3-3XYZ-W^3=0$. It contains your surface as an open subset, so to answer your question we might as well show that $S(\mathbb{Q})$ is dense in $S(\mathbb{R})$. Observe that $S$ has a singular rational point $P = (1:1:1:0)$. Since $P$ is singular, the intersection of $S$ with a plane $V$ that contains $P$ is a singular cubic $C_V$, with the rational point $P$ on it. It is a well-known and easy fact that on such curves, the rational points are dense (except if $C_V$ consists of three lines conjugate over $\mathbb{Q}$, but there are only finitely many $V$ for which this happens). 
Well, in order to approximate any real point $R \in S(\mathbb{R})$ to within distance $\epsilon$, we just pick a plane $V_0$ defined over $\mathbb{Q}$, which we may choose such as to be within distance $\epsilon$ to $R$. But then also the rational points on $C_{V_0}$, which lie dense in its real locus, lie within $\epsilon$ of $R$.