Consider a 3x3 cube, where the numbers in each cell are represented as a, b, c, d, e, f, g, h, and i.

The sum of all these numbers is 45 (a + b + c + d + e + f + g + h + i = 45), and the sum of each row, column, and diagonal must be equal to 15.

Let's assume that the sum of the numbers in each row is given by R, and the sum of the numbers in each column is given by C. We can then write the following equations:

a + b + c = R
d + e + f = R
g + h + i = R
a + d + g = C
b + e + h = C
c + f + i = C
a + e + i = C
c + e + g = C

We can then solve these equations to get:

a = (2C - R - e - i) / 2
b = (2C - R - d - f) / 2
c = (2C - R - e - g) / 2
h = (2C - R - b - e) / 2
g = (2C - R - h - i) / 2
f = (2C - R - d - b) / 2

Substituting these expressions into the equation for the sum of all the numbers in the cube, we get:

a + b + c + d + e + f + g + h + i = 45

Simplifying and rearranging terms, we get:

5C - 3R = 45

This equation implies that C and R must be odd and differ by a multiple of 3. However, this is not possible, since the sum of the numbers in each row, column, and diagonal of a 3x3 cube must be equal to 15, which is an odd number. Therefore, it is impossible to construct a semi-magic cube of order 3 using only cubes as the entries.