# N is a natural number such that when N3 divided by 9 ,it leaves remainder a.It can be concluded thata)a is a perfect square b) a is a perfect cube c) both of these d ) none of these

a is a perfect cube.
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
${N}^{3}=\left(3k{\right)}^{3}=9*3{k}^{3}$ which is divisible by 9. thus remainder a = 0
case 2: let N = 3k+1
${N}^{3}=\left(3k+1{\right)}^{3}=\left(3k{\right)}^{3}+{1}^{3}+3*\left(3k{\right)}^{2}*1+3*\left(3k\right)*{1}^{2}\phantom{\rule{0ex}{0ex}}=27{k}^{3}+27{k}^{2}+9k+1\phantom{\rule{0ex}{0ex}}=9\left(3{k}^{3}+3{k}^{2}+k\right)+1$
thus when ${N}^{3}$ is divided by 9, remainder a = 1, which is a perfect cube.
case 3: let N = 3k+2
${N}^{3}=\left(3k+2{\right)}^{3}=\left(3k{\right)}^{3}+{2}^{3}+3*\left(3k{\right)}^{2}*2+3*3k*{2}^{2}\phantom{\rule{0ex}{0ex}}=27{k}^{3}+8+54{k}^{2}+36k\phantom{\rule{0ex}{0ex}}=9\left(3{k}^{3}+6{k}^{2}+4k\right)+8$
thus when ${N}^{3}$ is divided by 9, remainder a = 8 which is a perfect cube.
hope this helps you.

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