<s>[INST] 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?
 [/INST] 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:
$$\left( \frac{251}{ \frac{1}{ \sqrt [3] {252} - 5 \sqrt [3] {2} } -10 \sqrt [3] {63} } - \frac {1} { \frac {251} { \sqrt [3] {252} +5 \sqrt [3] {2} } + 10 \sqrt [3] {63} } \right)^3=2000$$
</s>