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

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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

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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)

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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

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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

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if r = 7 a = 8q + 7a³ = (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.