# Show that the cube of a positive integer

Question:

Show that the cube of a positive integer of the form 6q + r,q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

Solution:

Let a be an arbitrary positive integer. Then, by Euclid’s division algorithm, corresponding to the positive integers ‘a’ and 6, there exist non-negative

integersg and r such that a = 6 q + r, where, 0 < r < 6

$a=6 q+r$, where, $0 \leq r<6$

$\Rightarrow \quad a^{3}=(6 q+r)^{3}=216 q^{3}+r^{3}+3 \cdot 6 q \cdot r(6 q+r)$

$\left[\because(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)\right]$

$\Rightarrow \quad a^{3}=\left(216 q^{3}+108 q^{2} r+18 q r^{2}\right)+r^{3}$ $\ldots$ (i)

where, $0 \leq r<6$

Case I When $r=0$, then putting $r=0$ in Eq. (i), we get

$a^{3}=216 q^{3}=6\left(36 q^{3}\right)=6 m$

where, $m=36 q^{3}$ is an integer.

Case II When $r=1$, then putting $r=1$ in Eq. (i), we get

$a^{3}=\left(216 q^{3}+108 q^{3}+18 q\right)+1=6\left(36 q^{3}+18 q^{3}+3 q\right)+1$

$\Rightarrow a^{3}=6 m+1$, where $m=\left(36 q^{3}+18 q^{3}+3 q\right)$ is an integer.

Case III When $r=2$, then putting $r=2$ in Eq. (i), we get

$a^{3}=\left(216 q^{3}+216 q^{2}+72 q\right)+8$

$a^{3}=\left(216 q^{3}+216 q^{2}+72 q+6\right)+2$

$\Rightarrow a^{3}=6\left(36 q^{3}+36 q^{2}+12 q+1\right)+2=6 m+2$

where, $\quad m=\left(36 q^{2}+36 q^{2}+12 q+1\right)$ is an integer.

Case IV When $r=3$, then putting $r=3$ in Eq. (i), we get

$a^{3}=\left(216 q^{3}+324 q^{2}+162 q\right)+27=\left(216 q^{3}+324 q^{2}+162 q+24\right)+3$

$=6\left(36 q^{3}+54 q^{2}+27 q+4\right)+3=6 m+3$

where, $m=\left(36 q^{2}+54 q^{2}+27 q+4\right)$ is an integer.

Case $V$ When $r=4$, then putting $r=4$ in Eq. (i), we get

$a^{3}=\left(216 q^{2}+432 q^{2}+288 q\right)+64$

$=6\left(36 q^{3}+72 q^{2}+48 q\right)+60+4$

$=a^{3} 6\left(36 q^{3}+72 q^{2}+48 q+10\right)+4=6 m+4$

where, $m=\left(36 q^{3}+72 q^{2}+48 q+10\right)$ is an integer.

Case VI When $r=5$, then putting $r=5$ in Eq. (i), we get

$a^{3}=\left(216 q^{3}+540 q^{2}+450 q\right)+125$

$\Rightarrow a^{3}=\left(216 q^{3}+540 q^{2}+450 q\right)+120+5$

$\Rightarrow a^{3}=6\left(36 q^{3}+90 q^{2}+75 q+20\right)+5$

$\Rightarrow a^{3}=6 m+5$

where, $m=\left(36 q^{3}+90 q^{2}+75 q+20\right)$ is an integer.

Hence, the cube of a positive integer of the form 6q + r,q is an integer and r = 0,1,2, 3, 4,5 is also of the forms 6m, 6m + 1, 6m + 2, 6m + 3,6m + 4 and 6m + 5 i.e.,6m + r.