use Euclid division lemma to show that the cube of any positive integer is of the form 9m, 9m+1, 9+8? find it in detail and simple way
ans me quick plz
Answers
a=x, b=3
0 < or = r < 3
case 1: if r=0,
x= 3q
cubing on both sides,
x^3= 27q^3 = 9 (9q^3) = 9m
where m= 3q^3
case 2: if r=1,
x= 3q+1
cubing on both sides,
x^3= (3q+1)^3
= 27q^3+27q^2+9q+1
= 9(3q^3+3q^2+q)+1
=9m+1
where m= 3q^3+3q^2+q
case 3: if r=2,
x= 3q+2
cubing on both sides,
x^3= (3q+2)
x^3= 27q^3+54q^2+36q+8
=9(3q^3q+6q^2+4q)+8
=9m+8
where m= 3q^3+6q^2+4q
Therefore, cube of any positive integer is in the form of 9m, 9m+1, 9m+8.
Step-by-step explanation:
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.
Hence, it is proved .
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