Question

Use euclid's division lemma to show that the cube of any positive integer is of the form 9m,9m+1 or 9m+8 taking a=9q+r

Answer 1

By euclid's division lemma a=bq+r, let b=3, where 'a' is any positive integer.

let r=0,1

when r=0,

a=3q+r

a=3q

cubing both sides...

a³=(3q)³

a³=27q³

a³=9(3q³)

let 3q³=m

then, a=9m

and

when r=1

a=3q+1

cubing both sides..

a³=(3q+1)³

{(a+b)³=a³+b³+3ab(a+b)}

=> a³= (3q)³+(1)³+(3)(3q)(1){3q+1)

=> a³= 27q³+1+9q(3q+1)

=> a³= 27q³+1+27q²+9q

=> a³= 27q³+27q²+9q+1

=> a³= 9(3q³+3q²+q)+1

let 3q³+3q²+q=m

=> a³=9m+1

same as given u can do the last part also taking r=2

$HOPE \: IT \: HELPS \: UH!$

Answer 2

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 .

THANKS.