show that any cube numbers is in the form of 4m,4m+1 or 4m+3
Answers
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To prove :-
Any cube numbers is in the form of 4m,4m+1 or 4m+3
Proof :-
Let's consider 'a' be any positive integer and b = 4.
Then by using Euclid Division Lemma,
a = bq + r [ 0 ≤ r < b ]
a = 3q + r [ 0 ≤ r < 4 ]
So, the values of r be :
r = 0, r = 1, r = 2, r = 3
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If, we put r = 0
a = 4q + 0
a = 4q
On cubing both sides,
▪️ a³ = (4q)³
▪️ a³ = 4 (16q³)
▪️ a³ = 9m [16q³ = m as integer]
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If we put , r = 1
a = 4q + 1
On cubing both sides ;
▪️ a³ = (4q + 1)³
▪️ a³ = 64q³ + 1³ + 3 × 4q × 1 ( 4q + 1 )
▪️ a³ = 64q³ + 1 + 48q² + 12q
▪️ a³ = 4 ( 16q³ + 12q² + 3q ) + 1
▪️ a³ = 4m + 1 [ Take m as some integer ]
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If we put, r = 2,
a = 4q + 2
On cubing both sides :
▪️ a³ = (4q + 2)³
▪️ a³ = 64q³ + 2³ + 3 * 4q * 2 ( 4q + 2 )
▪️ a³ = 64q³ + 8 + 96q² + 48q
▪️ a³ = 4 ( 16q³ + 2 + 24q² + 12q )
▪️ a³ = 4m [Consider m as some integer]
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If we put , r = 3
a = 4q + 3
On cubing both the sides;
▪️ a³ = (4q + 3)³
▪️ a³ = 64q³ + 27 + 3 * 4q * 3 (4q + 3)
▪️ a³ = 64q³ + 24 + 3 + 144q² + 108q
▪️ a³ = 4 (16q³ + 36q² + 27q + 6) + 3
▪️ a³ = 4m + 3 [Consider m as some integer]
Hence proved that the cube of any positive integer is in the form of 4m, 4m+1 or 4m+3.