Question

Show 4m, 4m+1 and 4m+3 as cube of positive integers

Answer 1

Step-by-step explanation:

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.

__________________

Identity used ;

$\because$ ( a + b )³ = a³ + b³ + 3ab ( a + b ) .

Hence, it is solved.