N is a natural number such that when N^{3} divided by 9 ,it leaves remainder a.It can be concluded that

a)a is a perfect square b) a is a perfect cube c) both of these d ) none of these

by the Euclid's Lemma any number can be represented as either of 3k , 3k+1 or 3k+2 form

case 1: let N = 3k

${N}^{3}=(3k{)}^{3}=9*3{k}^{3}$ which is divisible by 9. thus remainder a = 0

case 2: let N = 3k+1

${N}^{3}=(3k+1{)}^{3}=(3k{)}^{3}+{1}^{3}+3*(3k{)}^{2}*1+3*(3k)*{1}^{2}\phantom{\rule{0ex}{0ex}}=27{k}^{3}+27{k}^{2}+9k+1\phantom{\rule{0ex}{0ex}}=9(3{k}^{3}+3{k}^{2}+k)+1$

thus when ${N}^{3}$ is divided by 9, remainder a = 1, which is a perfect cube.

case 3: let N = 3k+2

${N}^{3}=(3k+2{)}^{3}=(3k{)}^{3}+{2}^{3}+3*(3k{)}^{2}*2+3*3k*{2}^{2}\phantom{\rule{0ex}{0ex}}=27{k}^{3}+8+54{k}^{2}+36k\phantom{\rule{0ex}{0ex}}=9(3{k}^{3}+6{k}^{2}+4k)+8$

thus when ${N}^{3}$ is divided by 9, remainder a = 8 which is a perfect cube.

hope this helps you.

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