Question

Use Euclid's division lemma to show that the cube of any positive
9m, 9m + 1 or 9m+8.
​

Answer 1

Step-by-step explanation:

let a be any positive integer

then a can have

a =3q, 3q+1, 3q+2

we prove that cube of these can be rewritten as 9m, 9m+1, 9m+8

(3q)^3 =27q^3 =9×3q^3 =9m

where m= 3q^3

(3q+1)^3 =27q^3 +1 +3×3q×1(3q+1)

=27q^3 +27q^2+9q+1

=9(3q^3 +3q^2+q) +1

=9m+1

(3q+2)^3 =27q^3 +8+3×3q×2 (3q+2)

=27q^3+8+8q(3q+2)

=27q^3+54q^2+36q+8

=9(3q^3+6q^2+4q)+8

=9m+8

where m = (3q^3 +6q^2 +4q)

Hence proved

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 .

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THANKS.