For, if possible, let D be cube.
Now C is also cube;
for it is fourth from the unit. [IX. 8]
And, as C is to D, so is B to C;
therefore B also has to C the ratio which a cube has to a cube.
And C is cube;
therefore B is also cube. [VIII. 25]
And since, as the unit is to A, so is A to B, and the unit measures A according to the units in it,
therefore A also measures B according to the units in itself;
therefore A by multiplying itself has made the cube number B.
But, if a number by multiplying itself make a cube number, it is also itself cube. [IX. 6]
Therefore A is also cube: which is contrary to the hypothesis.
Therefore D is not cube.
Similarly we can prove that neither is any other of the numbers cube except the fourth from the unit and those which leave out two.
Q. E. D.