First, simplify $\frac{251}{ \frac{1}{\sqrt[3]{252}-5\sqrt[3]{2}}-10\sqrt[3]{63}}$, then simplify $\frac{1}{\frac{251}{\sqrt[3]{252}+5\sqrt[3]{2}}+10\sqrt[3]{63}}$.

Let $x=\sqrt[3]{252},y=5\sqrt[3]2$, just to simplify things a bit. Note that $x^3-y^3=2,x^3+y^3=502$ and $xy=10\sqrt[3]{63}$. $$\frac{251}{\frac1{x-y}-xy}=\frac{251}{\frac{x^2+xy+y^2}{x^3-y^3}-xy}=\frac{251}{\frac{x^2+xy+y^2}2-xy}=\frac{251}{\frac{x^2-xy+y^2}2}=\frac{502(x+y)}{x^3+y^3}=x+y$$$$\frac1{\frac{251}{x+y}+xy}=\frac{251}{\frac{x^2-xy+y^2}{x^3+y^3}-xy}=\frac1{\frac{x^2-xy+y^2}2+xy}=\frac1{\frac{x^2+xy+y^2}2}=\frac{2(x-y)}{x^3-y^3}=x-y$$Therefore, $a=(x+y)+(x-y)=2x\implies a^3=\boxed{2016}$, which is quite a surprising answer.