use euclid division Lemma to show that cube of any positive integer is either of the form 9 M ( 9 mand + 1) or (9m + 8)
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Hey!
______________________________________________________________
Let a = 3q + r
a = 3q
•Cube both sides
(a)³ = (3q)³
= (a) = 9q³
Let q³ = m
= 9m
_______________________________
Let a = 3q + 1
a = 3q + 1
•Cube both sides
(a)³ = (3q + 1)³
(a)³ = 27q³ + 27q² + 9q + 1
•Take 9 as common :-
(a)³ = 9 (3q³ + 3q² + q) + 1
•Let (3q³ + 3q² + q) = m
(a)³ = 9m + 1
_______________________________
Let a = 3q + 2
Cube both sides
(a)³ = (3q + 2)³
(a)³ = 27q³ + 54q² + 36q + 8
•Take 9 common
(a)³ = 9 (3q³ + 6q² + 4q) + 8
•Let (3q³ + 6q² + 4q) = m
(a)³ = 9m + 8
_____________________________________________________________________________________________
Regards :)
Cybary
Be Brainly :)
______________________________________________________________
Let a = 3q + r
a = 3q
•Cube both sides
(a)³ = (3q)³
= (a) = 9q³
Let q³ = m
= 9m
_______________________________
Let a = 3q + 1
a = 3q + 1
•Cube both sides
(a)³ = (3q + 1)³
(a)³ = 27q³ + 27q² + 9q + 1
•Take 9 as common :-
(a)³ = 9 (3q³ + 3q² + q) + 1
•Let (3q³ + 3q² + q) = m
(a)³ = 9m + 1
_______________________________
Let a = 3q + 2
Cube both sides
(a)³ = (3q + 2)³
(a)³ = 27q³ + 54q² + 36q + 8
•Take 9 common
(a)³ = 9 (3q³ + 6q² + 4q) + 8
•Let (3q³ + 6q² + 4q) = m
(a)³ = 9m + 8
_____________________________________________________________________________________________
Regards :)
Cybary
Be Brainly :)
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