Irish Mathematical Olympiad

Let $a^3 = 252$ and $b^3 = 250$, then $ab = \sqrt[3]{252^3/250} = 10\sqrt[3]{63}$. Using that $251 = \frac{a^3+b^3}{2}$ and $1 = \frac{a^3-b^3}{2}$, the given expression inside the bracket becomes

$$\frac{\frac{a^3+b^3}{2}}{\frac{a^3-b^3}{2} - ab} + \frac{\frac{a^3-b^3}{2}}{\frac{a^3+b^3}{2} + ab}$$

The second term is obtained from the first by replacing $b$ with $-b$. Therefore, we consider only one of the two terms. The first term is equal to

$$\frac{a^3 + b^3}{\frac{a^3 - b^3}{a-b} - 2ab} = \frac{a^3 + b^3}{a^2 + ab + b^2 - 2ab} = \frac{a^3 + b^3}{a^2 - ab + b^2} = a + b.$$

Hence, the second term is equal to $a-b$ and the given number is equal to

$$((a+b) + (a-b))^3 = (2a)^3 = 8 \cdot 252 = 2016.$$