use Euclid division lemma to show that the cube ofany positive integer is the form 9q or 9q+1 or 9q+8 where q is some integer
Answers
Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
∴ r = 0,1,2 .
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 = (3q +1) ³
a = 27q ³+ 27q ² + 9q + 1
a = 9(3q ³ + 3q ² + q) + 1
a = 9m + 1 [ Where m = 3q³ + 3q² + q ) .
Case 3: When a = 3q + 2,
a = (3q +2) ³
a = 27q³ + 54q² + 36q + 8
a = 9(3q³ + 6q² + 4q) + 8
a = 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.
Answer:
let q be any positive integer. then,it is of the form 3q,3q+1,3q+2
9m,9m+1 or 9m+8
(3q)cube=27qcube=9(3qcube)
=9m. where m=3qcube
3q+1cube=(3qcube)+3(3q)squar*1+3(3q)*1square+1
=27qcube+54qsquar+36q+8
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