# 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 = $$3m^3 + 6m^2 + 4m$$