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.

Answer 1

let a be any positive integer which when divided by 3 gives r as remainder and q as quotient
so by Euclid division lemma
a=bq+r
a=3q+r
0<r<3
r=0,1,2
case 1 ,when r=0
a=3q+0
a=3q
a^3= (3q)^3
a^3=9(3q)^3
a^3=9m for some integer m

case 2 ,when r=1
a=3q+1
a^3=(3q+1)^3
a^3=27q^3 +1 +27q^2+9q
a^3=9(3q^3+3q^2+q)+1
a^3=9m+1 for some integer m

case 3, when r=2
a=3q+2
a^3=(3q+2)^3
a^3=27q^3+8+54q^2+36q
a^3=9(3q^3+6q^2+4q)+8
a^3=9m+8 for some integer m

here you answer buddy :)

Answer 2

Answer:

It is possible.

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,

⇒ a = ( 3q )³

⇒ a = 9( 3q³ )

⇒ a = 9m [ Where m = 3q³ ]

→ Case 2: When a = 3q + 1,

⇒ a = (3q +1)³

⇒ a = 27q³ + 27q² + 9q + 1

⇒ a = 9(3q³ + 3q² + 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 = (3q³ + 6q² + 4q) ]

Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Thanks ..!!