Rational solutions to x^3 + y^3 + z^3 - 3xyz = 1 I can show that there infinitely many solutions to this equation. Is it possible that the set
of rational solutions is dense?
 A: I think this surface has a rational parameterization in terms of (a,b), given by:
$x = (1 + a + a^2)^2/(9 (3 + a (6 + (-1 + a)^2 a)) b^2) + ((-2 + (-2 + a) a) b)/(1 + a + a^2)$ 
$y = (1 + a + a^2)^2/(9 (3 + a (6 + (-1 + a)^2 a)) b^2) + (b + 2 a b)/(1 + a + a^2)$
$z = (1 + a + a^2)^2/(9 (3 + a (6 + (-1 + a)^2 a)) b^2) + (b - a^2 b)/(1 + a + a^2)$
For rational (a,b), this should give you a dense set of rational points...

Let me explain where this parameterization comes from, so it'll be clear that this indeed shows that rational solutions are dense.
Consider the (linear, rational) change of variables: 
$x=p+r$,
$y=q+r$,
$z=r−p−q$. 
The equation then simplifies to: $p^2+pq+q^2=1/(9r)$. Since the quadratic is psd, for real points we need $r>0$. Each slice (for fixed $r$) is just an ellipse. If $r$ is the square of a rational number (dense on $\mathbb{R}$), we can parameterize all the rational solutions for that slice. Taking the union over real slices (for all $r>0$), we're done. 
A: 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 $R$.
