65. In a tetrahedron, two opposite edges are perpendicular, their lengths are $a$ and $b$, and the distance between them is $c$. A cube is inscribed in the tetrahedron, four of its edges are perpendicular to these two edges of the tetrahedron, and on each face of the tetrahedron lie exactly two vertices of the cube. Find the edge of the cube.
65. Let the common perpendicular to the given edges be divided by the cube into segments $y, x$ and $z, y+x+z=c(x-$ edge of the cube, $y$ adjacent to the edge $a$). The faces of the cube, parallel to the given edges, intersect the tetrahedron in two rectangles, the sides of the first being $\frac{x+z}{c} a, \frac{y b}{c}, y$ and the second - $\frac{z}{c} a, \frac{x+y}{c} b$, with the smaller sides of these rectangles equal to the edge of the cube, i.e., $\frac{y}{c} b=x$, $\frac{z}{c} a=x$, from which

$$
y=\frac{c x}{b}, \quad z=\frac{c x}{a} \text { and } x=\frac{a b c}{a b+b c+c a} .
$$