Dear Student

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,
x^2= (3q)^3= 27q^3= 9(3q^3)= 9m;
where m = 3q^3
Case (ii):When r = 1, then,
x^3= (3q+1)^3= (3q)^3+1^3+3×3q×1(3q+1) = 27q^3+1+27q^2+9q
Taking 9 as common factor, we get,
x^3= 9(3q^3+3q^2+q)+1
Putting(3𝑞^3+ 3𝑞^2+ 𝑞)= m, we get,
Putting (3q^3+3q^2+q) = m, we get ,
x^3= 9m+1
Case (iii): When r = 2, then,
x^3= (3q+2)^3= (3q)^3+2^3+3×3q×2(3q+2) = 27q^3+54q^2+36q+8
Taking 9 as common factor, we get,

x^3=9(3q^3+6q^2+4q)+8
Putting (3q^3+6q^2+4q) = m,
we get ,
x^3= 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.

Thanks