Use Euclid's division lemma to show that the cube of any positive integer is of the form $9 \mathrm{~m}, 9 \mathrm{~m}$ $+1$ or $9 m+8$.

Solution:

Any positive integer is of the form

$3 q, 3 q+1,3 q+2$

Case1: Let, $\mathrm{n}=3 \mathrm{q}$

Cube of this will be

$n^{3}=27 q^{3}$

$n^{3}=9\left(3 q^{3}\right)$

So, $\mathrm{n}^{3}=9 \mathrm{~m}$, where $\mathrm{m}=3 \mathrm{q}^{3}$

Case2: $\mathrm{n}=3 \mathrm{q}+1$

So, $n^{3}=(3 q+1)^{3}$

$\mathrm{n}^{3}=27 \mathrm{q}^{3}+1+27 \mathrm{q}^{2}+9 \mathrm{q}$

$=9\left(3 q^{3}+3 q^{2}+q\right)+1$

$\mathrm{n}^{3}=9 \mathrm{~m}+1$, where $\mathrm{m}=3 \mathrm{q}^{3}+3 \mathrm{q}^{2}+\mathrm{q}$

Case3: $\mathrm{n}=3 \mathrm{q}+2$

So, $n^{3}=(3 q+2)^{3}$

$=27 q^{3}+54 q^{2}+36 q+8$

$=9\left(3 q^{3}+6 q^{2}+4 q\right)+8$

$n^{3}=9 m+8$, where $m=3 q^{3}+6 q^{2}+4 q$

So, it means cube of positive integer is of the form $9 m, 9 m+1$ and $9 m+8$.