Let x = length of edge of the cube

**Sphere inscribed in a cube**

The diameter of the sphere is equal to edge of the cube.

$V_1 = \frac{1}{6}\pi x^3$

**Sphere circumscribing a cube**

From the figure above (f = face diagonal of the cube)

$f^2 = x^2 + x^2$

$f^2 = 2x^2$

s = space diagonal of the cube

$s^2 = x^2 + f^2$

$s^2 = x^2 + 2x^2$

$s^2 = 3x^2$

The diameter D of the sphere is equal to the space diagonal s of the cube. Thus,

$D^2 = 3x^2$

$D = x\sqrt{3}$

$V_2 = \frac{1}{6}\pi D^3$

$V_2 = \frac{1}{6}\pi (x\sqrt{3})^3$

$V_2 = \frac{1}{6}\pi x^3 \sqrt{27}$

**Comparison of V**_{1} and V_{2}

From the above results, it is clear that the volume of circumscribing sphere is square root of 27 times larger than the volume of inscribed sphere.

$V_2 = \sqrt{27} \, V_1$ *answer*