Question

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

Answer 1

Let x be any positive integer then it's form 6q+r

squareing

(6q+r)³=(6q)³+(1)³+3(6q)²(1)+3(6q)(1)²

=216q³+1+108q²+18q

= 6q[36q²+18q+1]+1

= 6q+1 [36q²+18q+1=1]

=6q+r

=6m+r [ Let q=m]