Let a and b be two positive integers, and a > b
a = (b × q) + r where q and r are positive integers and 0 ≤ r < b
Let b = (If 9 is multiplied by 3 a perfect cube number is obtained) a = 3q + r where 0 ≤ r < 3
(i) if r = 0, a = q (ii) if r = 1, a =3 q + 1 (iii) if r = 2, a = 3q + 2
Consider, cubes of these
Case (i) a = 3q
a3 = (3q)3 = q3 = 9(3q3) = 9m where m = 3q3 and 'm' is an integer.
Case (ii) a = 3q + 1
a3 = (q+1)3 [(a+b)3 = a3 + b3 + 3a2b + 3ab2 ]
= 27q3 + 1 + q2 + 9q = 27q3 + 27q2 + 9q + 1
= 9(3q3 + 3q2 + q) + 1 = 9m + 1
where m = 3q3 + 3q2 + q and ' m ' is an integer.
Case (iii) a = 3q + 2
a3 = (3q + 2)3 = q3 + 8 + 54q2+36q
= 27q3 - 54q2 + 36q + 8 = 9(3q3 + q2 + 4q) + 8
9m + 8, where m = 3q3 + 6q2 + 4q and m is an integer.
∴ Cube of any positive integer is either of the form 9m,9m+1 or 9m+8 for some integer m.