1.

Use Euclids divisionLemma to show that the cube of any positive integer is either of the form `9m , 9m+1`or, `9m+8`for some integer `m`.

Answer» Let a be any positive integer .
Let `" " b = 3`
`therefore " " a = 3q +r`
where , `0 le r lt 3` i.e. , r = 0 , 1, 2,
(i) When r = 0 , then
a =3q
`implies " " a^(3) = (3q)^(3) = 27q^(3) = 9(3q^(3)) = 9m`
where , `m = 3q^(3)` is an integer .
(ii) When , r =1 , then
a = 3q +1
`implies " " a^(3) = (3q +1) ^(3) = 27q^(3) + 27q^(2) + 9q + 1 = 9(3q^(3) + 3q^(2) + 1) + 1 = 9m +1`
where , ` m = 3q^(3) + 3q^(2) +1 ` is an integer .
(iii) When r = 2 , then
a = 3q + 2
`implies " " a^(3) = (3q +1)^(3)`
=` 27q^(3) + 54q^(2) + 36 q + 8 = 9(3q^(3) + 6q^(2) + 4q) + 8 = 9m + 8`
where , `m = 3q^(3) + 6q^(2) + 4q` is an integer .
Hence , the cube of any positive integer is of the form 9m or 9m +1 or 9m + 8. Hence Proved .


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