013 Insciribed and circumscribed sphere about a cube - volume comparison

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 V₁ and V₂
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