use euclid's divison lemma to show that cube of every positive integer is of the the form 9m or 9m+1 or 9m +8
Answers
B=3 ,r<3
Ie remainder =0,1,2
Case 1
When r =0 a=bq
Squaring,=a2=(3q)2
=9q2
=3(3q)2
=3m
Case 2
When r=1 a=bq+1
Squaring,(3q+1)2
9q2+6q+1
=3(3q2+2q)+1
=3m+1
Case 3
When r = 2 a=3q+squaring =(3q+2)2
=9q2+12q+4
=9q2+12q+3+1
3(3q2+4q+1)+1
3m+1
Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
Therefore, every number can be represented as these three forms. There are three cases.
Case 1: When a = 3q,
Where m is an integer such that m =
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.