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Choose one projective line from each of two different pairs. They are skew lines by the preceding disjointness result. After a projective linear coordinate change call themFor $A\in L$, $B\in M$, their joining line meets the cubic surface at $A,B$ and a third point. This defines a rational map $L\times M⇢S$ on the Zariski-open set where that third intersection is distinct and the joining line is not contained in $S$.

More explicitly, because $F$ vanishes on both projective lines,The third point is $[VA-UB]$. Both polynomials $U,V$ are nonzero: if either vanished identically, one of the two projective lines would lie in the singular locus. Thus $UV\ne 0$ supplies a nonempty Zariski-open set.

Conversely, a point $P=[x_0:x_1:x_2:x_3]$ outside $L\cup M$ lies on the unique joining line whose endpoints have coordinates $[x_0:x_1]$ on $L$ and $[x_2:x_3]$ on $M$. This is a rational inverse on the same generic locus. The rational parametrization of a cubic surface from two skew lines givesThe latter contains the affine plane as a dense open subset, as does ${\mathbb{P}}^2$. Hence $S$ is birational to ${\mathbb{P}}^2$.

The parametric equation of a straight line through $a$ with unit direction $\hat{t}$ isIf the two lines meet, then $a_1+s{\hat{t}}_1=a_2+t{\hat{t}}_2$ for some real $s,t$, so $a_1-a_2$ lies in the linear span of their direction vectors. Its dot product with their cross product is consequently zero:This is not sufficient without excluding parallel lines. For instance $a_1=(0,0,0)$, $a_2=(0,1,0)$ and ${\hat{t}}_1={\hat{t}}_2=(1,0,0)$ satisfy it, but the lines are distinct and never meet. For nonparallel lines the condition is sufficient, because the two directions span the plane perpendicular to their nonzero cross product.

For nonparallel skew lines, put $n=({\hat{t}}_1\times {\hat{t}}_2)/|{\hat{t}}_1\times {\hat{t}}_2|$. Every joining vector is $a_1-a_2+s{\hat{t}}_1-t{\hat{t}}_2$, with fixed component $(a_1-a_2)\cdot n$ along $n$. Its length is at least the absolute value of that component. Its perpendicular component can be canceled by a unique choice of $s,t$, since the two independent directions span $n^{\perp }$. This lower bound is attained, proving the distance between skew lines:

For two skew lines $L,M$ on a smooth cubic surface, their joining line meets the surface at a third point generically. In coordinates with $L=\{x_2=x_3=0\}$ and $M=\{x_0=x_1=0\}$, write $F(\alpha A+\beta B)=\alpha \beta (\alpha U+\beta V)$. The third point is $[VA-UB]$ whenever $UV\ne 0$. Neither $U$ nor $V$ is identically zero, because that would make one of the two lines singular on the surface. The inverse records $[x_0:x_1]$ and $[x_2:x_3]$ of a general surface point. Thus the surface is birational to ${\mathbb{P}}^1\times {\mathbb{P}}^1$, hence to ${\mathbb{P}}^2$.