Let x be any positive integer and y = 3.

By Euclid’s division algorithm, then,

x = 3q + r, where q ≥ 0 and r = 0, 1, 2, as r ≥ 0 and r < 3.

Therefore, putting the value of r, we get,

x = 3q

or

x = 3q + 1

or

x = 3q + 2

Now, by taking the cube of all the three above expressions, we get,

Case (i): When r = 0, then,

x²= (3q)³ = 27q³= 9(3q³)= 9m; where m = 3q³

Case (ii): When r = 1, then,

x³ = (3q+1)³ = (3q)³ +1³+3×3q×1(3q+1) = 27q³+1+27q²+9q

Taking 9 as common factor, we get,

x³ = 9(3q³+3q²+q)+1

Putting = m, we get,

Putting (3q³+3q²⁺q) = m, we get ,

x³ = 9m+1

Case (iii): When r = 2, then,

x³ = (3q+2)³= (3q)³+2³+3×3q×2(3q+2) = 27q³+54q²+36q+8

Taking 9 as common factor, we get,

x³=9(3q³+6q²+4q)+8

Putting (3q³+6q²+4q) = m, we get ,

x³ = 9m+8

Therefore, from all the three cases explained above, it is proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.