show that cube of integer can be expressed in one of the forms 9k,9k+1, 9k+8
Answers
Answered by
0
Answer:
Step-by-step explanation:
Let a be any positive integer and b = 3
a = 3q + r, ( where q ≥ 0 and 0 ≤ r < 3 )
Case 1: When a = 3q,
Where m is an integer such that m = 3q.
Case 2: When a = 3q + 1,
a^3 = (3q +1)^3
a^3 = 27q^3 + 27q^2 + 9q + 1
a^3 = 9(3q^3 + 3q^2 + q) + 1
a^3 = 9m + 1
Where m is an integer such that m = (3q^3 + 3q^2 + q)
Case 3: When a = 3q + 2,
a^3 = (3q +2)^3
a^3 = 27q^3 + 54q^2 + 36q + 8
a^3 = 9(3q^3 + 6q^2 + 4q) + 8
a^3 = 9m + 8
Where m is an integer such that m = (3q^3 + 6q^2 + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.
Similar questions