Question

use euclid's division Lemma to show that the cube of any positive integer is of the form 9m,9m+1,9m+8

Answer 1

Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
a = 3q or 3q + 1 or 3q + 2
Therefore, every number can be represented as these three forms.
There are three cases.
Case 1: When a = 3q,
a3 = (3q)3 = 27q3 = 9(3q3)= 9m
Where m is an integer such that m = 3q3
Case 2: When a = 3q + 1,
a3 = (3q +1)3
a3 = 27q3 + 27q2 + 9q + 1
a3 = 9(3q3 + 3q2 + q) + 1
a3 = 9m + 1
Where m is an integer such that m = (3q3 + 3q2 + q)
Case 3: When a = 3q + 2,
a3 = (3q +2)3
a3 = 27q3 + 54q2 + 36q + 8
a3 = 9(3q3 + 6q2 + 4q) + 8
a3 = 9m + 8
Where m is an integer such that m = (3q3 + 6q2 + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or
9m + 8.

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Answer 2

Hey Friends!!

Here is your answer↓

Let a be the any positive integers
and b=3.

By Euclid's Division lemma
a=bq+r. , 0≤r<b
0≤r<3
Hence,r=0,1,2.

Case 1.
Taking r=0
a=3q+0
=(3q)³
=27q³
=9(3q³)
=9m ( where m=3q³).

Case 2.
Taking r=1.
a=3q+1
=(3q+1)³
=27q³+1+27q²+9q.
=9(3q³+3q²+q)+1
=9m+1 (where m=3q³+3q²+q).

Case 3.
Taking r=2.
a=3q+2
=(3q+2)³
=27q³+8+54q²+72q.
=9(3q³+6q²+8q)+8
=9m+8 (where m=3q³+6q²+8q).

Hence,it is proved that some integers are 9m,9m+1,9m+2.

☺☺☺Hope it is helpful for you ✌✌✌.