Math, asked by guddashanu5831, 1 year ago

Use division algorithm to show that the cube of any positive integer is of the form 9m,9m+1,9m+8.

Answers

Answered by bharath0719
7

Use ED lemma has

a=bq+r ,a=3q+r,

a³=3q³a=27q³,9(3q³)

,9m where m=3q³

Is r=1,a=3q+1

a³=(3q+1³)=(3q)³+(3q)²(1)+3(3q)(1)²+1³

=27q³+9q²+9q

=9(3q³+q²+q) it gives 9m+1

Then a³=(3q+2)

=(3q³)+3q(3q²)(2)+3(3q)(2)²

=27q³+54q²+36q+8

=9(3q³+6q²+4q)+8

=9m+8 where m = (3q³+6q+4q)

from the above we find that the cube of any positive integer is of the form 9m, 9m+1,9m+8

Answered by Anonymous
7

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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