Question

Show that the cubeof any postive integer is of the form. (4m)or 4m+1 or 4m+3 some integer m
Hint n 4q,4q+1,4q+2,4q+3

Answer 1

$\huge{\bold{Answer:}}$

$\bf \underline{Step-by-step explanation:}$

Let 'a' be any positive integer and b = 4.

Using Euclid Division Lemma,

a = bq + r           [ 0 ≤ r < b ]

⇒ a = 4q + r      [ 0 ≤ r < 4 ]

Now, possible value of r :

r = 0, r = 1, r = 2, r = 3

$\huge{\underline{\bold{CASE \: I :}}}$

If we take, r = 0

⇒ a = 4q + 0

⇒ a = 4q

On cubing both sides,

⇒ a³ = (4q)³

⇒ a³ = 4 (16q³)

⇒ a³ = 4m         [16q³ = m as integer]

$\huge{\underline{\bold{CASE \: II :}}}$

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 ]

$\huge{\underline{\bold{CASE \: III :}}}$

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]

$\huge{\underline{\bold{CASE \: IV :}}}$

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 ;

( a + b )³ = a³ + b³ + 3ab ( a + b )

Answer 2

Answer:

Step-by-step explanation:

Let "a" be any positive integer and b = 4.

By Euclid Division Lemma, the positive value of r = 0, 1, 2, 3 & 4.

Now, The Positive value of "a" be 4m+0, 4m+1, 4m+2, 4m+3.

Case |,

a = 4m+0

Cubing on both the sides,

a³ = ( 4m +0)³

=> a³ = 64m³

=> a³ = 4(16m³)

Let a³ = x & (16m³) =m. (Where m is any positive integer).

=> X = 4m.

Case ||,

a = 4m+1

a³ = (4m+1)³

=> a³ = 64m³ + 1 + 3*4m*1 ( 4m+1)

=> a³ = 64m³ + 1 + 12m*4m + 12m*1

=> a³ = 64m³ + 48m² + 12m +1

=> a³ = 4 ( 16m³ + 12m² + 3m ) +1

Let's a = X & (16m³ + 12m² + 3m) = m.

=> X = 4m +1.

Case |||,

a³ = ( 4m +2)³

=> a³ = 64m³ + 8 + 3*4m*2 ( 4m +2)

=> a³ = 64m³ + 8 + 24m*4m + 24m*2

=> a³ = 64m³ + 96m² + 48m + 8

=> a³ = 4 ( 16m³ + 24m² + 12m +2 )

Let's a³ = X & ( 16m³ + 24m² + 12m +2 )= m.

=> X = 4m.

Case |V,

a³ = (4m+3)³

=> a³ = 64m³ + 27 + 36m ( 4m +3)

=> a³ = 64m³ + 27 + 144m² + 108m

=> a³ = 4 ( 16m³ + 36m² + 27m + 6) +3

Let's a³ = X & ( 16m³ + 36m² + 27m + 6) = m.

=> X = 4m +3.

Hence,

The cube of any positive integer is of the form 4m , 4m+1 and 4m+3.

Formula Used :-

(a+b)³ = a³ + b³ + 3ab(a+b).

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