prove that cube of any natural number is of the form 4q or 4q+ 1 or 4q + 3
Answers
Case1: (4q)3 = 64q3 = 4 (16q3) where m = 16q3
Case 2: (4q + 1)3 = (4q)3 + 1+ 3 (16q2) + 3 (4q) = 64q3+ 48q2+ 12q +1 = 4q (16q2+ 12q + 3) + 1 where m = q (16q2 + 12q+ 3) for 4m+ 1
Case 3: (4q+ 2)3 = (4q)3+ 8+ 3 (4q)2 (2) + 3 (4q) (4) = 64q3+ 96q2+ 48q + 8 = 4 (16q3+ 24q2+ 12 + 2) where m= (16q3+ 24q2+ 12 + 2) for 4m
Case 4: (4q+3)3= (4q)3 + 27 + 3 (4q)2 (3) + 3 (4q) (9)= 64q3+ 36q2+ 108q+ 24+3 = 4 (16q3+ 9q2+ 27q+ 6) + 3 where m =(16q3+ 9q2+ 27q+ 6) for 4m+3
Hence cube of any positive integer is of the form 4m or 4m+1 or 4m+3
hope it helps you!!
Step-by-step explanation:
Note :- I'm taking m as some integer.
Let 'a' be any positive integer and b = 4.
Using Euclid Division Lemma,
a = bq + r [ 0 ≤ r < b ]
⇒ a = 3q + r [ 0 ≤ r < 4 ]
Now, possible value of r :
r = 0, r = 1, r = 2, r = 3
CASE 1:
If we take, r = 0
⇒ a = 4q + 0
⇒ a = 4q
On cubing both sides,
⇒ a³ = (4q)³
⇒ a³ = 4 (16q³)
⇒ a³ = 9m [16q³ = m as integer]
CASE 2:
If we take, 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 ]
CASE 3:
If we take 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 [Take m as some integer]
CASE 4 :
If we take, 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 [Take m as some integer]
Hence, the cube of any positive integer is in the form of 4m, 4m+1 or 4m+3.
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Identity used ;
∵ ( a + b )³ = a³ + b³ + 3ab ( a + b ) .
Hence, it is solved.