If (x²+y²)/(x²-y²) + (x²-y²)/(x²+y²) = k
Write the following in terms of x
(x⁸+y⁸)/(x⁸-y⁸) + (x⁸-y⁸⁾/(x⁸+y⁸⁾

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Given}, \\ \frac{{\mathrm{x}}^2+{\mathrm{y}}^2}{{\mathrm{x}}^2-{\mathrm{y}}^2}+\frac{{\mathrm{x}}^2-{\mathrm{y}}^2}{{\mathrm{x}}^2+{\mathrm{y}}^2}=\mathrm{k} \\ \Rightarrow \frac{{\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}^2+{\left({\mathrm{x}}^2-{\mathrm{y}}^2\right)}^2}{\left({\mathrm{x}}^2-{\mathrm{y}}^2\right)\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}=\mathrm{k} \\ \Rightarrow \frac{{\mathrm{x}}^4+{\mathrm{y}}^4+2{\mathrm{x}}^2{\mathrm{y}}^2+{\mathrm{x}}^4+{\mathrm{y}}^4-2{\mathrm{x}}^2{\mathrm{y}}^2}{{\mathrm{x}}^4-{\mathrm{y}}^4}=\mathrm{k} \\ \Rightarrow \frac{2.\left({\mathrm{x}}^4+{\mathrm{y}}^4\right)}{{\mathrm{x}}^4-{\mathrm{y}}^4}=\mathrm{k} \\ \Rightarrow \frac{\left({\mathrm{x}}^4+{\mathrm{y}}^4\right)}{\left({\mathrm{x}}^4-{\mathrm{y}}^4\right)}=\frac{\mathrm{k}}{2}--- \left(1\right) \\ \frac{\left({\mathrm{x}}^4-{\mathrm{y}}^4\right)}{\left({\mathrm{x}}^4+{\mathrm{y}}^4\right)}=\frac{2}{\mathrm{k} } ----\left(2\right) \\ \mathrm{rationalizing} \mathrm{and} \mathrm{adding} 1 \mathrm{and} 2 , \mathrm{we} \mathrm{get} \\ \frac{\left({\mathrm{x}}^4+{\mathrm{y}}^4\right)}{\left({\mathrm{x}}^4-{\mathrm{y}}^4\right)}\frac{\left({\mathrm{x}}^4+{\mathrm{y}}^4\right)}{\left({\mathrm{x}}^4+{\mathrm{y}}^4\right)}+\frac{\left({\mathrm{x}}^4-{\mathrm{y}}^4\right)}{\left({\mathrm{x}}^4+{\mathrm{y}}^4\right)}\frac{\left({\mathrm{x}}^4-{\mathrm{y}}^4\right)}{\left({\mathrm{x}}^4-{\mathrm{y}}^4\right)}=\frac{\mathrm{k}}{2}+\frac{2}{\mathrm{k}} \\ \mathrm{taking} \mathrm{L}.\mathrm{H}.\mathrm{S}, \mathrm{we} \mathrm{get} \\ \frac{\left({\mathrm{x}}^8+{\mathrm{y}}^8+2{\mathrm{x}}^4{\mathrm{y}}^4\right)}{\left({\mathrm{x}}^8-{\mathrm{y}}^8\right)}+\frac{\left({\mathrm{x}}^8+{\mathrm{y}}^8-2{\mathrm{x}}^4{\mathrm{y}}^4\right)}{\left({\mathrm{x}}^8-{\mathrm{y}}^8\right)} \\ \Rightarrow \mathrm{cancelling} 2{\mathrm{x}}^4{\mathrm{y}}^4, \mathrm{we} \mathrm{get} \\ 2.\frac{\left({\mathrm{x}}^8+{\mathrm{y}}^8\right)}{\left({\mathrm{x}}^8-{\mathrm{y}}^8\right)}=\frac{{\mathrm{k}}^2+4}{2\mathrm{k}} \\ \frac{\left({\mathrm{x}}^8+{\mathrm{y}}^8\right)}{\left({\mathrm{x}}^8-{\mathrm{y}}^8\right)}=\frac{{\mathrm{k}}^2+4}{4\mathrm{k}} \\ \mathrm{similarly}, \frac{\left({\mathrm{x}}^8-{\mathrm{y}}^8\right)}{\left({\mathrm{x}}^8+{\mathrm{y}}^8\right)}=\frac{4\mathrm{k}}{{\mathrm{k}}^2+4} \\ \mathrm{Hence} \\ \frac{\left({\mathbf{x}}^{\mathbf{8}}\mathbf{+}{\mathbf{y}}^{\mathbf{8}}\right)}{\left({\mathbf{x}}^{\mathbf{8}}\mathbf{-}{\mathbf{y}}^{\mathbf{8}}\right)}\mathbf{+}\frac{\left({\mathbf{x}}^{\mathbf{8}}\mathbf{-}{\mathbf{y}}^{\mathbf{8}}\right)}{\left({\mathbf{x}}^{\mathbf{8}}\mathbf{+}{\mathbf{y}}^{\mathbf{8}}\right)}\mathbf{=}\frac{{\mathbf{k}}^{\mathbf{2}}\mathbf{+}\mathbf{4}}{\mathbf{4}\mathbf{k}}\mathbf{+}\frac{\mathbf{4}\mathbf{k}}{{\mathbf{k}}^{\mathbf{2}}\mathbf{+}\mathbf{4}}\mathbf{=}\frac{{\left({\mathbf{k}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\right)}^{\mathbf{2}}\mathbf{+}\mathbf{16}{\mathbf{k}}^{\mathbf{2}}}{\mathbf{4}\mathbf{k}\left({\mathbf{k}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\right)}\mathbf{=}\frac{{\mathbf{k}}^{\mathbf{4}}\mathbf{+}\mathbf{24}{\mathbf{k}}^{\mathbf{2}}\mathbf{+}\mathbf{16}}{\mathbf{4}{\mathbf{k}}^{\mathbf{3}}\mathbf{+}\mathbf{16}\mathbf{k}} \end{gathered}$$

Hope this information will clear your doubts about this topic.

If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
Regards