Question

Please solve this que..

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Answer 1

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Let 'a' be any positive integer,

$a = 4q+r$

$r = 0, 1, 2, 3$

Case (i), r = 0

$= (4q)^3$

$= 64q^3$

$= 4(16q^3)$

It is in the form '4m' where where ' m = 16q³ '

Case (ii), r = 1

$= (4q+1)^3$

$= 64q^3 + 48q^2 + 12q + 1$

$= 4(16q^3 + 12q^2 +3) + 1$

It is of the form '4m+1' where ' m = 16q³ + 12q²+ 3 '

Case (iii), r = 2

$= (4q+2)^3$

$= 64q^3 + 96q^2 + 48q + 8$

$= 4(16q^3 + 24q^2 + 12q + 2)$

It is of the form '4m' where ' m = 16q³ + 24q² + 12q + 2 '

Case (iv), r = 3

$= (4q+3)^3$

$= 4(16q^3 + 36q^2 + 27q + 6) + 3$

It is of the form '4m+3' where ' m = 16q³ + 36q² + 27q + 6 '

Hence, the cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for any positive integer 'm'

Answer 2

Answer:

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