Q. 14.0( 29 Votes )

Show that the cub

Answer :

By Euclid’s division algorithm, b = a × q + r, 0 ≤ r < a


Here, b is any positive integer .


Let a be an arbitrary positive integer. Then corresponding to the positive integers ‘a’ and 6, there exist non-negative integers q and r such that


a = 6q + r where 0 ≤ r < 6


Cubing both the sides using (a + b)3 = a3 + b3 + 3ab(a + b)


a3 = (6q + r)3 = 216q3 + r3 + 3 × 6q × r(6q + r)


a3 = (216q3 + 108q2r + 18qr2) + r3


where 0 ≤ r < 6


Case I When r = 0 we get a3 = 216q3


a3 = 6(36q3) a3 = 6m + 0 where m = 36q3 is an integer


Case II When r = 1 we get


a3 = (216q3 + 108q2 + 18q) + 1


a3 = 6(36q3 + 18q2 + 3q) + 1


a3 = 6m + 1, where m = (36q3 + 18q2 + 3q) is an integer.


Case III When r = 2 we get


a3 = 6(36q3 + 36q2 + 12q + 1) + 2


a3 = 6m + 2


where m = (36q3 + 36q2 + 12q + 1) is an integer.


Case IV When r = 3 we get


a3 = (216q3 + 324q2 + 162q) + 27


a3 = (216q3 + 324q2 + 162q + 24) + 3


a3 = 6(36q3 + 54q2 + 27q + 4) + 3


a3 = 6m + 3


where m = (36q2 + 54q2 + 27q + 4) is an integer.


Case V When r = 4 we get


a3 = (216q2 + 432q2 + 288q) + 64


a3 = 6(36q3 + 72q2 + 48q) + 60 + 4


a3 = 6(36q3 + 72q2 + 48q + 10) + 4


a3 = 6m + 4


where m = (36q3 + 72q2 + 48q + 10) is an integer.


Case VI When r = 5 we get


a3 = (216q3 + 540q2 + 450q) + 120 + 5


a3 = 6(36q3 + 90q2 + 75q + 20) + 5


a3 = 6m + 5


where m = (36q3 + 90q2 + 75q + 20) + 5


Hence, the cube of a positive integer of the form 6q + r, q, is an integer and r = 0,1,2,3,4,5 is also of the forms 6m, 6m + 1, 6m + 2, 6m + 3,6m + 4and 6m + 5i.e., 6m + r.


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