Use Euclid's division lemma to show that cube of any positive integer is either in form of 9m, 9m+1 or 9m+8
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Let the positive integer be a
therefore , it will be in form of 3q , 3q + 1 , 3q+2
a*3 = ( 3q )^3
a^3 = 27 q^3
a^3 = 9 × 3q^3 where 3q^3 = m
a^3 = 9m.
if ,
a^3 = (3q +1 )^3
a^3 = 27q^3 + 1 + 9q ( 3q + 1 )
a ^3 = 27q^3 + 1 + 27q^2 + 9q
a^3 = 9 ( 3q^3 + 3q^2 + q ) + 1
where 3q^3+ 3q^2 + q = m
a^3 = 9m + 1
if ,
a^3 = ( 3q + 2 )^3
a^3 = 27q^3 + 8 + 18q ( 3q+ 2)
a^3 = 27q^3 + 8 + 54q^2 + 36q
a^3 = 9 ( 3q^3 + 6q^2 + 4q ) + 8
where 3q^3 +6q^2 + 4q = m
a^3 = 9m+ 8
Hence proved....
Hope it helps..
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therefore , it will be in form of 3q , 3q + 1 , 3q+2
a*3 = ( 3q )^3
a^3 = 27 q^3
a^3 = 9 × 3q^3 where 3q^3 = m
a^3 = 9m.
if ,
a^3 = (3q +1 )^3
a^3 = 27q^3 + 1 + 9q ( 3q + 1 )
a ^3 = 27q^3 + 1 + 27q^2 + 9q
a^3 = 9 ( 3q^3 + 3q^2 + q ) + 1
where 3q^3+ 3q^2 + q = m
a^3 = 9m + 1
if ,
a^3 = ( 3q + 2 )^3
a^3 = 27q^3 + 8 + 18q ( 3q+ 2)
a^3 = 27q^3 + 8 + 54q^2 + 36q
a^3 = 9 ( 3q^3 + 6q^2 + 4q ) + 8
where 3q^3 +6q^2 + 4q = m
a^3 = 9m+ 8
Hence proved....
Hope it helps..
Mark as Brainliest....
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