a) As the sum of $a$, $b$, and $c$ is divisible by $6$, and is therefore even, there must be either $0$ or $2$ odd numbers among the three. If we had $2$ odd numbers, the sum of the squares $a^2 + b^2 + c^2$ would give a remainder of $0 + 1 + 1 = 2$ when dividing by $4$. But this is not possible, since the sum is divisible by $36$, and therefore also by $4$. So, all the numbers $a$, $b$, $c$ are even. Hence, all the numbers $a^3$, $b^3$, $c^3$ are divisible by $8$, and so is their sum.

b) If $a = 8$, $b = c = 2$, then all of the premises are fulfilled: $8 + 2 + 2 = 12$ is divisible by $6$ and $8^2 + 2^2 + 2^2 = 72$ is divisible by $36$. But $8^3 + 2^3 + 2^3 = 528$ is not divisible by $9$, and therefore, it is not divisible by $27$.