Math, asked by asishpandupaoil5, 1 year ago

use division algorithm to show that the cube of any positive integer is of the form 9m or 9m+1 or 9m+8

Answers

Answered by creamiepie
18
We know,
For every positive integer a and b there always exists another two integers q and r such that a=bq+r and 0<=r<b.


Taking b=3
a=3q+0 or
a=3q+1 or
a=3q+2
Now,
a^3=(3q)^3
=27q^3
=9×3q^3
=9m[Taking 3q^3=m]

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
=9m+1[Taking 3q^3+3q^2+q=m]

a^3=(3q+2)^3
=(3q)^3 + 3×(3q)^2×2 + 3×3q×2^2+2^3
=27q^3 + 54q^2 +36q+8
=9(3q^3+6q^2+4q)+8
=9m+8[Taking 3q^3+6q^2+4q=m]



proved....


hope it helps u....
Answered by Anonymous
10

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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