v={d}{-8π}

Question

Its related to Volume transformation Please Prove that

Answer 1

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.