Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m or 9m + 1 or 9m + 8.

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 = \(3m^3 + 6m^2 + 4m\)