# 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?

$$\begin{eqnarray*} \sqrt[3]\frac{(\sqrt{a-1} - \sqrt{a})^5}{(\sqrt{a-1} + \sqrt{a})} + \sqrt[3]\frac{(\sqrt{a-1} + \sqrt{a})^5}{(\sqrt{a} - \sqrt{a-1})} = \frac{-(\sqrt{a-1} -\sqrt{a})^2 + (\sqrt{a-1} -\sqrt{a})^2}{\sqrt[3]{(\sqrt{a} - \sqrt{a-1})(\sqrt{a} + \sqrt{a-1})}} \\ = \frac{-(a-1+a-2 \sqrt{a(a-1)})+ a-1+a+2 \sqrt{a(a-1)}}{\sqrt[3]{a-(a-1)}}=4\sqrt{a(a-1)} \end{eqnarray*}$$
• Hello, did you solve it correctly? Because the answer given to me is $4\sqrt{a(a-1)}$. Maybe you had some typos? Jan 2, 2021 at 17:24