Question

use Euclid division lemma to show that the cube of any positive integer is of the form 9m or 9m+1 or 9m+8​

Answer 1

Answer:

let a be any positive integer and b = 3

Step-by-step explanation:

a = bq+ r where q> = 0 and 0<= r <b

so I take b = 3

then a= 3q+ r where q>= 0 and 0<= r< 3

there are three cases

1 : case

when a = 3q

then a^ 3 = (3q)^ 3 =27q^3

9 (3q^2)= 9 m where m = 3q^2

2 : case

when a = 3q + 1

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

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

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

where 3q ^ 3+ 3q ^2+ 1 = m

hence 9m+ 1

3 : case

when a = 3q + 2 = a^ 3 = (3q + 2 )^3

= 27 q^3 + 8 + 54 q^2 + 18 q

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

= 9 m + 8

where m = 3q ^ 3 + 6 q ^2 + 2q

Done ....

Answer 2

Answer:

hi tammana sorry for rude behavior

username : adarsh_thakur_05

Step-by-step explanation:

Let x be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.

So, we have the following cases :

Case I : When x = 3q.

then, x3 = (3q)3 = 27q3 = 9 (3q3) = 9m, where m = 3q3.

Case II : When x = 3q + 1

then, x3 = (3q + 1)3

= 27q3 + 27q2 + 9q + 1

= 9 q (3q2 + 3q + 1) + 1

= 9m + 1, where m = q (3q2 + 3q + 1)

Case III. When x = 3q + 2

then, x3 = (3q + 2)3

= 27 q3 + 54q2 + 36q + 8

= 9q (3q2 + 6q + 4) + 8

= 9 m + 8, where m = q (3q2 + 6q + 4)

Hence, x3 is either of the form 9 m or 9 m + 1 or, 9 m + 8.