Is there an 'interesting' way to derive this expression? So I was asked to prove the following term is equal to $2016$: 
$$ \left( \frac{251}{ \frac{1}{ \sqrt [3] {252} - 5 \sqrt [3] {2} } -10 \sqrt [3] {63} } + \frac {1} { \frac {251} { \sqrt [3] {252} +5 \sqrt [3] {2} } + 10 \sqrt [3] {63} } \right)^3 $$
For the record, I know how to solve the question; defining $a^3=252 $ and $b^3 = 250 $ simplifies the expression so it can be solved. 
But I was hoping (and this may be too much to hope for) that there was some reason why this radical expression simplifies so nicely, and would give a straightforward way of deriving the problem. So is there anything special to this expression?

This is related to your approach, but possible easier. Hard to tell since you didn't give your argument.
$$\begin{align}
f(x,y)&=\frac{\frac{1}{2}(x^3+y^3)}{\frac{\frac{1}{2}(x^3-y^3)}{x-y}-xy}\\
&=\frac{x^3+y^3}{\frac{x^3-y^3}{x-y}-2xy}\\
&=\frac{x^3+y^3}{x^2+xy+y^2-2xy}\\
&=\frac{x^3+y^3}{x^2-xy+y^2}\\
&=x+y
\end{align}$$
Then $f(x,y)+f(x,-y)=2x$. 
In the case of $x=\sqrt[3]{252},y=\sqrt[3]{250}=5\sqrt[3]{2}$, $\frac{x^3+y^3}{2}=251, \frac{x^3-y^3}{2}=1, xy=10\sqrt[3]{63}$.
This also lets you see that:
$$\left( \frac{251}{ \frac{1}{ \sqrt [3] {252} - 5 \sqrt [3] {2} } -10 \sqrt [3] {63} } - \frac {1} { \frac {251} { \sqrt [3] {252} +5 \sqrt [3] {2} } + 10 \sqrt [3] {63} } \right)^3=2000$$
