Math, asked by nandupetaj35p5xbat, 1 year ago

use e.d.l show cube of any positive integer is of form 9m,9m+1,9m+8 (as please only send last 3 steps plz plz)

Answers

Answered by Riyaaaaaa
0
Hope this helps you.. :p
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Answered by Anonymous
1

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


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