Question 589243
The longest distance between vertices in a cube is {{{sqrt(3)}}} times the side of the cube.
 
REASON:
The diagonal of one of the faces of a cube is the hypotenuse of a right triangle that has for legs the sides of that face of the cube, of length s. So the length of the diagonal of a face is
{{{sqrt(s^2+s^2)=sqrt(2s^2)=sqrt(2)*s}}}
That diagonal and the adjacent edge from another face for a right triangle, with leg lengths {{{s}}} and {{{sqrt(2)*s}}}. The hypotenuse is the diagonal of the cube, and its length is
{{{sqrt(s^2+(sqrt(2)*s)^2)=sqrt(s^2+2s^2)=sqrt(3s^2)=sqrt(3)*s}}}
 
If the cube is incribed in a spere of diameter 9 inches, the cube has a 9 inch diagonal and
{{{sqrt(3)*s=9}}} --> {{{s=9/sqrt(3)=3sqrt(3)}}}
The volume of a cube with a side length of {{{s=9/sqrt(3)=3sqrt(3)}}} is
{{{V=(3sqrt(3))^3=3^3*(sqrt(3))^3=27*3sqrt(3)=81sqrt(3)}}} cubic inches
(approximately 140.3 cubic inches).