Let m be any positive integer. Then it is of the form 3m, 3m + 1 or 3m + 2.<\/p>\n
Now, we have to prove that the cube of these can be rewritten in the form 9q, 9q + 1 or 9q + 8.<\/p>\n
Now,<\/p>\n
\\((3m)^3\\) = \\(27m^3\\) = \\(9(m^3)\\)<\/p>\n
= 9q, where q = \\(3m^3\\)<\/p>\n
\\((3m + 1)^3\\) = \\((3m)^3\\) + \\(3(3m)^2\\).1 + 3(3m).\\(1^2\\) + 1<\/p>\n
= \\(27m^3\\) + \\(27m^2\\) + 9m +1<\/p>\n
= \\(9(3m^3 + 3m^2 + m)\\) + 1<\/p>\n
= 9q + 1, where 1 = \\(3m^3 + 3m^2 + m\\)<\/p>\n
and \\((3m + 1)^3\\) = \\((3m)^3\\) + \\(3(3m)^2\\).2 + 3(3m).\\(2^2\\) + 8<\/p>\n
= \\(27m^3\\) + \\(54m^2\\) + 36m + 8<\/p>\n
= \\(9(3m^3 + 6m^2 + 4m)\\) + 8<\/p>\n
= 9q + 8, where q = \\(3m^3 + 6m^2 + 4m\\)<\/p>\n","protected":false},"excerpt":{"rendered":"
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 …<\/p>\n