Answer:

Let US consider a and b where a be any positive number and b is equal to 3.

According to Euclid’s Division Lemma

a = bq + r

where r is greater than or equal to zero and LESS than b (0 ≤ r < b)

a = 3q + r

so r is an INTEGER greater than or equal to 0 and less than 3.

Hence r can be either 0, 1 or 2.

Case 1: When r = 0, the equation becomes

a = 3q

Cubing both the sides

a³= (3q)³

a³ = 27 q³

a3 = 9 (3q3)

a³ = 9m

where m = 3q³

Case 2: When r = 1, the equation becomes

a = 3q + 1

Cubing both the sides

a³ = (3q + 1)³

a³ = (3q)³+ 13 + 3 × 3q × 1(3q + 1)

a³ = 27q³+ 1 + 9q × (3q + 1)

a³ = 27q³+ 1 + 27q² + 9q

a³= 27q³ + 27q² + 9q + 1

a³ = 9 ( 3q³ + 3q² + q) + 1

a³ = 9m + 1

Where m = ( 3q³ + 3q²+ q)

Case 3: When r = 2, the equation becomes

a = 3q + 2

Cubing both the sides

a³ = (3q + 2)³

a³= (3q)³ + 23 + 3 × 3q × 2 (3q + 1)

a³= 27q³ + 8 + 54q² + 36q

a³ = 27q³+ 54q² + 36q + 8

a³= 9 (3q³ + 6q² + 4q) + 8

a³ = 9m + 8

Where m = (3q3 + 6q2 + 4q)therefore a can be any of the form 9m or 9m + 1 or, 9m + 8.