Question

Show that the cube of any positive integer is of the form 3p, 3 p 1 or 3p 8 by use eculids division lemma

Answer 1

/* There is a mistake in the question. It may be like this*/

$\red { Show \:that \:the \:cube \:of\:any }$

$\red { positive \:integer \:is \:of \:the \:form }$

$\red{ 9p, 9p+1\:Or \:9p+8 ,by \:use \: Euclid's}$

$Let \: \pink{m} \: be \: any \: positive \: integer\\then \:it \:is \:in \:the \:form \: of \:3m, 3m+1 \:or \\3m+2$

$Now, we \:have \: to \:prove \:that \:the \:cube \\of \:each \:of \:these \:can \:be \: rewritten \:in \\the \:form \:of \:9q,9q+1,\:or \:9q+8.$

$Now, (3m)^{3}\\ = 27m^{3} \\= 9(3m^{3})\\9q, \:where \: p = 3m^{3}$

$(3m+1)^{3} = 27m^{3}+27m^{2}+9m+1 \\= 9(3m^{3}+3m^{2}+m) + 1 \\= 9p+1 , \\where \:p = 3m^{3}+3m^{2}+m$

$(3m+2)^{3} = 27m^{3} + 54m^{2} + 36m + 8 \\=9(3m^{3} + 6m^{2} + 4m )+8 \\= 9p+8\\where \: p = 3m^{3} + 6m^{2} + 4m$

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