Question

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


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Answer 1

$\huge\bf\pink{\mid{\overline{\underline{Your\: Answer}}}\mid}$

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

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CASEI:

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]

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CASEII:

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

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CASEIII:

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

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CASEIV:

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 )

Answer 2

Let a be the +ve integer and b = 4.

$\text{\underline{Then,}}$

$\text{\underline{By\:Euclid's\:algorithm,}}$

${a = 4q + r}$ for some integer q ≥ 0 and r = 0, 1, 2, 3

$\text{\underline{Because,}}$ 0 ≤ r < 4.


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$\text{\underline{So,}}$

$\sf{a = 4q\:or\:(4q + 1)}$ or $\sf{(4q + 2)\:or\:(4q + 3)}$

$\sf{(4q)^{3}}$

$\sf{= 64 {q}^{3}}$

$\sf{= 4(16 {q}^{3} )}$

$\sf{= 4m,}$ $\text{where\:m\:is\:some\:integer.}$


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$\sf{(4q + 1) ^{3}}$

$\sf{= 64 {q}^{3} + 48 {q}^{2} + 12q + 1}$

$\sf{ = 4(16 {q}^{3} + 12 {q}^{2} + 3) + 1}$

$\sf{ = 4m + 1,}$ $\text{where\:m\:is\:some\:integer.}$


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$\sf{(4q + 2) ^{3}}$

$\sf{ = 64 {q}^{3} + 96q ^{2} + 48q + 8}$

$\sf{ = 4(16 {q}^{3} + 24q ^{2} + 12q + 2)}$

$\sf{ = 4m,}$ $\text{where\:m\:is\:some\:integer.}$


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$\sf{(4q + 3) ^{3}}$

$\sf{ = 64q^{3} + 144 {q}^{2} + 108q + 27}$

$\sf{= 4(16q ^{3} + 36q ^{2} + 27q + 6) + 3}$

$\sf{ = 4m + 3,}$ $\text{where\:m\:is\:some\:integer.}$


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$\text{\underline{Hence,}}$

The cube of any +ve integer is of the form, $\sf{4m,}$ $\sf{4m + 1}$or $\sf{4m + 3}$ for some integer m.