Problem 723

33. $A=(a-1)^{3}+a^{3}+(a+1)^{3}=3 a\left(a^{2}+2\right)$. If $a$ is divisible by 3, then $A$ is divisible by 9. If $a$ is not divisible by 3, then $a=$ $=3 t \pm 1 ; A=3(3 t \pm 1)\left(9 t^{2} \pm 6 t+3\right)=9(3 t \pm 1)\left(3 t^{2} \pm 2 t+\right.$ $+1)$ is divisible by 9.

Alternatively: $A=a^{3}+(a+1)^{3}+(a+2)^{3}=3 a\left(a^{2}+\right.$ $+5)+9 a^{2}+9=3 a\left(a^{2}-1+6\right)+9 a^{2}+9=3(a-1) a(a+$ $+1)+9 a^{2}+18 a+9$ is divisible by 9, since among three consecutive numbers $a-1, a, a+1$ one is necessarily divisible by 3.