Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Let a and b be two positive integergs such that a is greater than b; then :
a = bq + r;
where q and r are positive integers 0 ≤ r < b.
Taking b = 3, we get
a = 3q + r ; where 0 ≤ r < 3.
⇒ Different values of integer a are 3q, 3q + 1 or 3q + 2.
Cube of 3q = (3q)3
= 27q3 = 9(3q3) = 9m ;
where m is some integer.
Cube of 3q + 1 = (3q + 1)3
= (3q + 3(3q)2 × 1 + 3(3q) × 12+ l3
[∵ (a + b)3 = a3 + 3a2b + 3ab2 + 1]
= 27q3 + 27q2 + 9q + 1
= 9(3q3 + 3q2 + q) + 1
= 9m + 1; where m is some integer.
Cube of 3q + 2 = (3q + 2)3
= (3q)3 + 3(3q)2 × 2 + 3 × 3q × 22 + 23
= 27q3 + 54q2 + 36q + 8
= 9(3q3 + 6q2 + 4q) + 8 = 9m + 8; where m is some integer.
∴ Cube of any positive integer is of the form 9m or 9m + 1 or 9m + 8.
