Show that the cube of any positive integer is of the form 9m,9m+1 or 9m+8
Answers
ANSWER--
a=3q+r, 0_<r<3 , where q is a positive integer.
if,r=0
a=3q
a^3=(3q)^3
=9*(3q^3)
=9m
where m = 3q^3
now if , r=1
a=3q+1
a^3=(3q+1)^3
=(3q)^3+3.(3q^2).1+3.3q.(1)^2+(1)^3
=27q^3+27q^2+9q+1
=9(3q^3+3q^2+q)+1
m= 3q^3+3q^2+q
now if , r=2
a=3q+2
a^3=(3q+2)^3
=(3q)^3+3.(3q)^2.2+3.3q.(2)^3
=27q^3+54q^2+36q+8
=9(3q^3+6q^2+4q)+8
m=3q^3+6q^2+4q
Hope it helps,
thanks..
Answer:
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,
Now a=9q^3
Where m is an integer such that m = q^3
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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