Question

using Euclid's division lemma to show that the cube a positive integers is of the form of 9m ,9m+1,9m+8​

Answer 1

a = 9m + 1 [ Where m = 3q³ + 3q² + q ) . Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Answer 2

Answer:

We know that,

a=bq+r,let m be any positive integer.

Here b=3. so possible remainders are 3m,3m+1,3m+2. 0<r<3

Step-by-step explanation:

There are three cases ;

Case 1:  $n^3 =(3m)^3= 27m^3 = 3(9m^3)$

Case 2 : $n^3 = (3m+1)^3 = 27m^3+1+27m^2+9m$

                   $= 9(3m^3+3m^2+m)+1$

Case 3 : $n^3 = (3m+2)^3 = 27m^3+54m^2+36m+8$

                   $= 9(3m^3+6m^2+4m)+8$

By euclids division lemma , we had shown that the cube of any positive integer of the form  9m ,9m+1,9m+8​