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 that positive integer be a.

And b=3

So, according to Euclid lemma,

a=3q+r where r=0, 1, 2...(as we know that 0<r<b)

So, case 1-when r is 0,

a=3q

Cubing both sides,

a^3 = 27 q^3

a^2 = 9(3 q^3)

a^2 = 9m.......(3q^3 =m)

Case 2 - when r is 1,

a = 3q+1

Cubing both sides,

a^3 = (3q+1)^3

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

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

a^3 = 9m+1..{ (3 q^3 + 3 q ^2 + 1q) =m}

Similarly.... I hope you could solve the 3rd case..... :)

Hope my solution will help you

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.