use division to show that the cube of any positive integer is of the form 9m, 9m+1 or 9m+8
Answers
Answer:
Step-by-step explanation:
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 rather than or equal to 0 and less than equal to 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.
Hope it helps!
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