Math, asked by ap113552, 9 months ago

14a). Show that cube of any positive Integer will be in the form or 8m or
8m+1 or8m+3 or 8m+5 or 8m+7, where m is a whole number.​

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Answered by hanshraj26
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where is a whole number

by Arfakhurana 21.02.2019

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Panesarh989gmailcomAmbitious

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

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

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.

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