(a) Every perfect cube $k^{3}$ of a positive integer is special because we can write

$$k^{3}=k^{3} \frac{a^{3}+2 b^{3}}{a^{3}+2 b^{3}}=\frac{(k a)^{3}+2(k b)^{3}}{a^{3}+2 b^{3}}$$

for some positive integers $a, b$.
(b) Observe that $2014=2 \cdot 19 \cdot 53$. If 2014 is special, then we have,

$$x^{3}+2 y^{3}=2014\left(u^{3}+2 v^{3}\right)$$

for some positive integers $x, y, u, v$. We may assume that $x^{3}+2 y^{3}$ is minimal with this property. Now, we will use the fact that if 19 divides $x^{3}+2 y^{3}$, then it divides both $x$ and $y$. Indeed, if 19 does not divide $x$, then it does not divide $y$ too. The relation $x^{3} \equiv -2 y^{3} \pmod{19}$ implies $\left(x^{3}\right)^{6} \equiv \left(-2 y^{3}\right)^{6} \pmod{19}$. The latter congruence is equivalent to $x^{18} \equiv 2^{6} y^{18} \pmod{19}$. Now, according to Fermat's Little Theorem, we obtain $1 \equiv 2^{6} \pmod{19}$, that is 19 divides 63, not possible.
It follows $x=19 x_{1}, y=19 y_{1}$, for some positive integers $x_{1}$ and $y_{1}$. Replacing in (1) we get

$$19^{2}\left(x_{1}^{3}+2 y_{1}^{3}\right)=2 \cdot 53\left(u^{3}+2 v^{3}\right)$$

i.e. $19 \mid u^{3}+2 v^{3}$. It follows $u=19 u_{1}$ and $v=19 v_{1}$, and replacing in (2) we get

$$x_{1}^{3}+2 y_{1}^{3}=2014\left(u_{1}^{3}+2 v_{1}^{3}\right) .$$

Clearly, $x_{1}^{3}+2 y_{1}^{3}<x^{3}+2 y^{3}$, contradicting the minimality of $x^{3}+2 y^{3}$.