Question 1066287
Let a be the first number, b be the 2nd, c be the 3rd.

So a + b + c = 28 and  a*b*c = 512.

Normally this would not be enough, but we know that there is a progression.

Let  r be the common ratio between numbers in a geometric progression.

So the 2nd number can be written as:
b = r*a

And the 3rd number can be written as:

c = r^2*a

Go back to our original problem.

a + b + c  can be written as  a + ra + r^2*a

a*b*c  can be written as  a * (ra) * (r^2*a) =  r^3*a^3

so we have that two equations.

a+ra+r^2*a = 28
r^3*a^3 = 512

We can solve for ra  by taking the cube root of both sides in the 2nd equation.

(r*a)^3  = 512

(r*a) = 8

So a = 8/r

8/r +  8 + 8r = 28

8/r + 8r = 20

8(1/r + r) = 20

(r^2+1 / r) = 5/2

2(r^2+ 1) = 5r

2r^2 + 2 = 5r

2r^2 - 5r +2 = 0

we can rewrite as  (2r-1)(r-2) so  r = 1/2 or r = 2.

Let's try r = 2 first.

So  if  r =2 then a = 4

So we have 

4, 8, 16 which do satisfy the conditions. But are there other numbers?

Try  r = 1/2.   Then  a = 16.

So we have 16, 8 , 4. Ah, so it appears it is the same either way.

The numbers are 4, 8 , 16.