Form of cube of an integer

Question

Prove that the cube of any integer is of the form $9k \text {or} 9k\pm 1$

I tried $a=9k+r$ where $0\leq r<9$ Then I have to put value of r from r=0,1,...8 Is there any short method instead putting the value of r from 1 to 8

Answer 1

Hint: try writing $a = 3s + r$ where $0 \leq r < 3$. Now you only have 3 cases to consider rather than 9.