A tetrahedron in a cube

If $x$ is the length of the sides of the cube, then the volume of the cube is $V_c = x^3$ .

The sides of the tetrahedron are the diagonals of the cube faces, which would be $\sqrt{2}x$ . Since the volume of a tetrahedron with side lengths of s is $\frac{s^3}{6\sqrt{2}}$ , the volume of this tetrahedron is $V_t = \frac{(\sqrt{2}x)^3}{6\sqrt{2}} = \frac{x^3}{3}$ .

The ratio of the volume of the tetrahedron to the volume of the cube is $\frac{V_t}{V_c} = \frac{\frac{x^3}{3}}{x^3} = \boxed{\frac{1}{3}}$ .

Consider the other parts of the cube.4 congruent tetrahedron,the volume of one of them is one sixth of the cube,so $1-\frac16\times4=\frac13$