Simplify, $\sqrt[3]\frac{(\sqrt{a-1} - \sqrt{a})^5}{(\sqrt{a-1} + \sqrt{a})} + \sqrt[3]\frac{(\sqrt{a-1} + \sqrt{a})^5}{(\sqrt{a} - \sqrt{a-1})}$. 
Simplify, $\sqrt[3]\frac{(\sqrt{a-1} - \sqrt{a})^5}{(\sqrt{a-1} + \sqrt{a})} + \sqrt[3]\frac{(\sqrt{a-1} 
+ \sqrt{a})^5}{(\sqrt{a} - \sqrt{a-1})}$.

What I Tried: I thought of substituting $\sqrt{a - 1} = x$ , $\sqrt{a} = y$ .
This gives :- $$\rightarrow \sqrt[3]\frac{(x - y)^5}{(x + y)} + \sqrt[3]\frac{(x + y)^5}{(y - x)}$$
But I was not able to find any good factorisation for this. I even took some help from Wolfram Alpha and it gives me this :- 
Another thing I thought of was to substitute only $\sqrt{a} = x$. This would give :-
$$\rightarrow \sqrt[3]\frac{(\sqrt{(x + 1)(x - 1)} - x)^5}{(2x^2 - 1)} + \sqrt[3]\frac{(\sqrt{(x + 1)(x - 1)} + x)^5}{1}$$
This looks more or less simpler to work with, but unfortunately I could not get any ideas.
Can anyone help me?

\begin{eqnarray*}
\sqrt[3]\frac{(\sqrt{a-1} - \sqrt{a})^5}{(\sqrt{a-1} + \sqrt{a})} + \sqrt[3]\frac{(\sqrt{a-1} 
+ \sqrt{a})^5}{(\sqrt{a} - \sqrt{a-1})} = \frac{-(\sqrt{a-1} -\sqrt{a})^2 + (\sqrt{a-1} -\sqrt{a})^2}{\sqrt[3]{(\sqrt{a} - \sqrt{a-1})(\sqrt{a} + \sqrt{a-1})}} \\ 
= \frac{-(a-1+a-2 \sqrt{a(a-1)})+ a-1+a+2 \sqrt{a(a-1)}}{\sqrt[3]{a-(a-1)}}=4\sqrt{a(a-1)}
\end{eqnarray*}
