Solution :
Let m be any positive integer. Then it is of the form 3m, 3m + 1 or 3m + 2.
Now, we have to prove that the cube of these can be rewritten in the form 9q, 9q + 1 or 9q + 8.
Now,
\((3m)^3\) = \(27m^3\) = \(9(m^3)\)
= 9q, where q = \(3m^3\)
\((3m + 1)^3\) = \((3m)^3\) + \(3(3m)^2\).1 + 3(3m).\(1^2\) + 1
= \(27m^3\) + \(27m^2\) + 9m +1
= \(9(3m^3 + 3m^2 + m)\) + 1
= 9q + 1, where 1 = \(3m^3 + 3m^2 + m\)
and \((3m + 1)^3\) = \((3m)^3\) + \(3(3m)^2\).2 + 3(3m).\(2^2\) + 8
= \(27m^3\) + \(54m^2\) + 36m + 8
= \(9(3m^3 + 6m^2 + 4m)\) + 8
= 9q + 8, where q = \(3m^3 + 6m^2 + 4m\)
