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?

You can first get rid of the denominator in each fraction by multiplying above and below by the conjugate expression of the denominator. Then you will have 6th powers above, and you can get rid of the cubic roots.
