Question

Show that the cube of any positive integer is of the form 9m,9m+1 or9m+8for some integer m

Answer 1

hello,
Let x be any positive integer . Then, it is of the form  3q,3q+1,3q+2. So,we have the following cases.

CASE1:-When x=3q:in this case we have ,(HERE WE WILL CUBE ARE EVERY CASE)
 

x³=(3q)³=27q³
=>9(3q³)
=>9m
where m-is any positive integer and m=3q³.

CASE2:- When x=3q+1 here we have,

x³=(3q+1)³
=>27q³+27q²+9q+1
=>x³=9q(3q²+3q+1)+1
=>9m+1,where m=q(3q²+3q+1) and m is any positive integer.

CASE3:-

When x=3q+2;

x³=(3q+2)³
=>x³=27q³+54q²+36q+8
=>x³=9q(3q²+6q+4)+8
=>x³=9m+8, where m=q(3q²+6q+4) and m is any positive integer.
Hence,x³ is either of the form 9m,9m+1,9m+8.

HOPE IT HELPS YOU OUT!!

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

#BeBrainly.