Let a be any arbitrary positive integer such that
`a=6q+r, " ""where " 0 le r lt 6`
`a^(3) =(6q+r)^(3) =2 16q^(3) +r^(3) +3xx6q xxr(6q+r)`
`=216q^(3) +r^(3) +108q^(2) + 18qr^(2)`
`=(216q^(3) +108q^(2)r +18qr^(2))+r^(3),0le r lt 6` ...(1)
Case I : when r=0,
`therefore` `a^(3) =216q^(3) "[from (1))]"`
`=6(36q^(3))` =6m+0 , which is of the form 6m+r , where r=0
Case II When r=1,
`therefore` `a^(3) =216q^(3) +108q^(2)+18q+1` [from (1)]
`=6(36q^(3)+18q^(2)+3q)+1` ltbrge6m+1 , which is of the form 6m+ r where r=1
Case III : When r=2,
`therefore` `a ^(3) =(216q^(3)+ 216q^(2)+ 72q)+8` [from (1)]
`=6(36q^(3)+36q^(2)+12q +1)+2`
`=6m+2 `, which is of the form 6m+ r where r=2.
Case IV : When r=3,
`therefore` `a^(3) =(216q^(3) +324q^(2)+ 162q)+ 27` [from (1)]
`=6(36q^(2)+ 54q^(2) +27q+4)+3`
`=6m+3`, which is of the form 6m+ r where r=3
Case V : When r=4 ,
`therefore` `a^(3) =(216a^(3) +432q^(2)+ 288q)+64`
`=6(36q^(3) +72q^(2)+48q +10) +4`
`=6m+4` ,which is of the form 6m+ r , where r=4
Case VI : When r=5,
`therefore` `a^(3) =(216q^(3) +540q^(2) +450q)+125`
`=6(36q^(2) +90q^(2)+75q+20)+5`
`=6m+5`.which is of the form 6m+r , where r=5
Hence the cube of any positive integer of the form 6q+r,q is an integer and r=0,1,2,3,4 and 5 is also of the form 6m+r where r=0,1,2,3,4,5.
