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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 :- Here

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?

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2 Answers 2

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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.

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\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*}

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  • $\begingroup$ Hello, did you solve it correctly? Because the answer given to me is $4\sqrt{a(a-1)}$. Maybe you had some typos? $\endgroup$
    – Anonymous
    Jan 2, 2021 at 17:24
  • $\begingroup$ @Anonymous You are right ... I did not notice that the numerator of the first term is negative the denominator of the second term ... I have correct the solution now. $\endgroup$ Jan 2, 2021 at 17:54

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