b) It is enough to provide an example: $2016 = 1000 + 1000 + 8 + 8$.

a) Let us first note that $2016 = 2^5 \cdot 3^2 \cdot 7$. And consider remainders of $a^3, b^3, c^3$ modulo $7$. The remainders could equal to $0$ or $\pm 1$. Thus, if $2016 = x^3 + y^3 + z^3$ then at least one of the numbers is divisible by $7$. It is easy to see that it must be $7$, because $7^3 = 343 < 2016$, and the next number $14^3 = 2744 > 2016$. Thus two other numbers must satisfy the condition $y^3 + z^3 = 1673$.

Now, suppose that $y \le z$. Since $12^3 = 1728 > 1673$ then $z \le 11$. Moreover, $1673 = y^3 + z^3 \le 2z^3$, then $837 \le z^3$, and it means that $10 \le z$. To finish the solution we just need to check cases $z=10,11$.