Question

show that the cube of any positive interger will be in the form of 8m or 8m +1 or 8m +3or 8m+5 or 8m+7 where is a whole number ​

Answer 1

Let a be a positive integer.

According to Euclid division lemma,

a = bq + r where 0 ≤ r < b.

Let b = 8 ,

then, a = 8q + r

r can be 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 .

Let's consider r = 0

them a = 8q

Cubing on both sides we get,

a³ = (8q)³

= 512q³

= 8(64q³)

= 8m where m = 64q³

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If r = 1 ,

then a = 8q +1

a³ = (8q + 1)³

a³ = 512q³ + 1 + 3(8q)(8q+1)

= 512q³ + 1 + 24q(8q+1)

= 512q³ + 1 + 192q² + 24q

= 8( 64q³ + 24q² + 3q) + 1

= 8m + 1 where m = 64q³ + 24q² + 3q

===========================

If r = 2 ,

a = 8q + 2

a³ = (8q+2)³

= 512q³ + 8 + 48q(8q+2)

= 512q³ + 8 + 384q²+ 96q

= 512q³ + 384q²+ 96q + 8

= 8 ( 64q³ + 48q² + 12q + 1 )

= 8m

where m = 64q³ + 48q² + 12q+ 1

===========================

if r = 3

then a = 8q + 3

a = (8q+3)³

= 512q³+27+72q(8q+3)

= 8(64q³+ 3 + 72q²+ 27q ) + 3

= 8m+ 3

where m = 64q³ + 72q² + 27q+3)

=======================

If r = 4

a = 8q + 4

a³ = 512q³+ 64 + 768q² + 384q

a³ = 8( 64q³ + 8 + 96q² + 48q)

a³ = 8m

where m = 64q³ + 8 + 96q² + 48q

===========================

when r = 5

a = 8q + 5

a³ = (8q+5)³

= 512q³ + 960q² + 600q + 125

= 8 ( 64q³ + 120q² + 75q + 15) + 5

= 8m + 5

where m = 64q³ + 120q² + 75q + 15

===========================

if r = 6 .

then a = 8q + 6

a³ = ( 8q + 6)^3

= 512q³ + 1152q² + 864q + 216

= 8 ( 64q³ + 144q² + 108q + 27 )

= 8m

where m = 64q³ + 144q² + 108q + 27

===========================

if r = 7

a = 8q + 7

a³ = (8q + 7)³

= 512q³ + 343 + 1344q² + 1176 q

= 8( 64q³ + 168q² + 147q + 42) + 7

= 8m + 7

where m = 64q³ + 168q² + 147q + 42

===========================

Therefore, We proved that cube of a positive integer will be of the form 8m , 8m+ 1 , 8m+3 , 8m+5 , 8m+ 7 where m is a whole number.

Answer 2

$<b>$

Solution:-

Let "a" be any positive integer & b = 8.

By Euclid Division Lemma,

The positive value of "r" :- 0, 1, 2, 3, 4, 5, 6 & 7.

=> The Positive value of "a" be :-

(8m+0), (8m+1), (8m+2), (8m+3), (8m+4), (8m+5), (8m+6) & (8m+7).

Case 1st,

a = 8m + 0

Cubing on both the sides. we get,

$= > {a}^{3} = {(8m + 0)}^{3} \\ \\ = > {a}^{3} = 512 {m}^{3} \\ \\ = > {a}^{3} = 8(64 {m}^{3} )$

Now, Taking a^3 = x & 64m^3 = m(where m is any positive integer). we get,

=> x = 8m._____________(1)

Case 2nd,

a = 8m + 1

Cubing on both the sides. we get,

$= > {a}^{3} = {(8m + 1)}^{3} \\ \\ = > {a}^{3} = 512 {m}^{3} + 1 + 3 \times 8m \times 1(8m + 1) \\ \\ = > {a}^{3} = 512 {m}^{3} + 1 + 192 {m}^{2} + 24m \\ \\ = > {a}^{3} = 8(64 {m}^{3} + 24 {m}^{2} + 3m) + 1$

Taking a^3= x & (64m^3 + 24m^2 + 3m) = m( where m is any positive integer). we get,

=> x = 8m + 1. ___________(2).

Case 3rd,

a = 8m + 2

Cubing both the sides. we get,

$= > {a}^{3} = {(8m + 2)}^{3} \\ \\ = > {a}^{3} = 512 {m}^{3} + 8 + 3 \times 8m \times 2(8m + 2) \\ \\ = > {a}^{3} = 512 {m}^{3} + 8 + 384 {m}^{2} + 96m \\ \\ = > {a}^{3} = 8(64 {m}^{3} + 48 {m}^{2} + 12m + 1)$

Taking a^3 = x & (64m^3 + 48m^2 + 12m + 1) = m(where "m" is any positive integer). we get,

=> x = 8m. _____________(3).

Case 4,

a = 8m + 3

Cubing on both the sides. we get,

$= > {a}^{3} = {(8m + 3)}^{3} \\ \\ = > {a}^{3} = 512 {m}^{3} + 27 + 3 \times 8m \times 3(8m + 3) \\ \\ = > {a}^{3} = 512 {m}^{3} + 27 + 576 {m}^{2} + 216m \\ \\ = > {a}^{3} = 8(64 {m}^{3} + 72 {m}^{2} + 27m + 3) + 3$

Taking a^3 =x & (64m^3 + 72m^2 + 27m +3) = m(where m is any positive integer). we get,

=> x = 8m + 3.___________(4).

Case 5th,

a = 8m + 5

Cubing on both the sides. we get,

$= > {a}^{3} = {(8m + 5)}^{3} \\ \\ = > {a}^{3} = 512 {m}^{3} + 125 + 120m(8m + 5) \\ \\ = > {a}^{3} = 512 {m}^{3} + 125 + 960 {m}^{2} + 600m \\ \\ = > {a}^{3} = 8(64 {m}^{3} + 115 {m}^{2} + 75m + 16) + 3$

Taking a^3 =x and (64m^3 + 115m^2 + 75m + 16) = m( where m is any positive integer). we get,

=> x = 8m + 3.____________(5).

From eq 1, 2, 3, 4 & 5.

Cube of any positive integer is of the form 8m or (8m +1) or (8m +3).

Thanku!