I have to guess what this question means, given so little information.
The first thing I notice is the appearance of π, which is usually associated with circular measurements. The second thing I noticed is the word "volume". So perhaps we're dealing with the volume of a sphere. Perhaps d is the diameter of the sphere.
Then I see there is a minus sign -8π. So it occurred to me that the question might be: if the volume of a sphere is reduced by 8π, what is the reduction in size of the diameter? (If the volume was decreased by a factor of 8, the diameter would be halved because ½3=⅛.)
The volume of a sphere V=4πr3/3 where r is the radius. r=d/2, so the volume is 4π(d/2)3/3=4πd3/24=πd3/6.
V-8π=πd3/6-8π=π(d3/6-8)=π(d3-48)/6. So if this is the new volume v, we can write:
v=π(d3-48)/6. If this causes the diameter to change to D, then v=πD3/6.
So πD3/6=π(d3-48)/6, D3=d3-48, or D=∛(d3-48). Therefore D depends on what d was before the volume was removed. Another way of writing this is D3/d3=1-48/d3. So D/d=∛(1-48/d3). For this to make sense, d3 must be greater than 48. For example, let d=4 then D/d=∛(1-48/64)=∛¼, making D/d=0.63 approx, which means that D is 63% of d. Since d=4, D=2.52 approx.
The question asks for proof of something but doesn't state what is to be proved.