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Question Number 157223 by amin96 last updated on 21/Oct/21

Answered by Rasheed.Sindhi last updated on 21/Oct/21

$$\underset{-}{\frac{{\mathrm{x}}^2+{\mathrm{y}}^2}{{\mathrm{x}}^2-{\mathrm{y}}^2}+\frac{{\mathrm{x}}^2-{\mathrm{y}}^2}{{\mathrm{x}}^2+{\mathrm{y}}^2}=\mathrm{k};\frac{{\mathrm{x}}^8+{\mathrm{y}}^8}{{\mathrm{x}}^8-{\mathrm{y}}^8}+\frac{{\mathrm{x}}^8-{\mathrm{y}}^8}{{\mathrm{x}}^8+{\mathrm{y}}^8}=?}$$$$\frac{{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2+{({\mathrm{x}}^2-{\mathrm{y}}^2)}^2}{({\mathrm{x}}^2+{\mathrm{y}}^2)({\mathrm{x}}^2-{\mathrm{y}}^2)}=\mathrm{k}$$$$\frac{2({\mathrm{x}}^4+{\mathrm{y}}^4)}{{\mathrm{x}}^4-{\mathrm{y}}^4}=\mathrm{k}$$$$\frac{{\mathrm{x}}^4+{\mathrm{y}}^4}{{\mathrm{x}}^4-{\mathrm{y}}^4}=\frac{\mathrm{k}}{2}$$$$\frac{{\mathrm{x}}^4+{\mathrm{y}}^4}{{\mathrm{x}}^4-{\mathrm{y}}^4}+\frac{{\mathrm{x}}^4-{\mathrm{y}}^4}{{\mathrm{x}}^4+{\mathrm{y}}^4}=\frac{\mathrm{k}}{2}+\frac{2}{\mathrm{k}}$$$$\frac{{({\mathrm{x}}^4+{\mathrm{y}}^4)}^2+{({\mathrm{x}}^4-{\mathrm{y}}^4)}^2}{({\mathrm{x}}^4+{\mathrm{y}}^4)({\mathrm{x}}^4-{\mathrm{y}}^4)}=\frac{{\mathrm{k}}^2+4}{2\mathrm{k}}$$$$\frac{2({\mathrm{x}}^8+{\mathrm{y}}^8)}{{\mathrm{x}}^8-{\mathrm{y}}^8}=\frac{{\mathrm{k}}^2+4}{2\mathrm{k}}$$$$\frac{{\mathrm{x}}^8+{\mathrm{y}}^8}{{\mathrm{x}}^8-{\mathrm{y}}^8}=\frac{{\mathrm{k}}^2+4}{4\mathrm{k}}$$$$\frac{{\mathrm{x}}^8+{\mathrm{y}}^8}{{\mathrm{x}}^8-{\mathrm{y}}^8}+\frac{{\mathrm{x}}^8-{\mathrm{y}}^8}{{\mathrm{x}}^8+{\mathrm{y}}^8}=\frac{{\mathrm{k}}^2+4}{4\mathrm{k}}+\frac{4\mathrm{k}}{{\mathrm{k}}^2+4}$$$$=\frac{{({\mathrm{k}}^2+4)}^2+{\left(4\mathrm{k}\right)}^2}{2\mathrm{k}({\mathrm{k}}^2+4)}=\frac{{\mathrm{k}}^4+{8\mathrm{k}}^2+16+{16\mathrm{k}}^2}{4\mathrm{k}({\mathrm{k}}^2+4)}$$$$=\frac{{\mathrm{k}}^4+{24\mathrm{k}}^2+16}{4\mathrm{k}({\mathrm{k}}^2+4)}$$

Answered by som(math1967) last updated on 21/Oct/21

$$\frac{{(x^2+y^2)}^2+{(x^2-y^2)}^2}{(x^2-y^2)(x^2+y^2)}=k$$$$\Rightarrow \frac{x^4+y^4}{x^4-y^4}=\frac{k}{2}$$$$\Rightarrow \frac{x^4+y^4+x^4-y^4}{x^4+y^4-x^4+y^4}=\frac{k+2}{k-2}★$$$$\Rightarrow \frac{x^8}{y^8}={\left(\frac{k+2}{k-2}\right)}^2$$$$\Rightarrow \frac{x^8+y^8}{x^8-y^8}=\frac{2(k^2+4)}{8k}★$$$$\therefore \frac{x^8+y^8}{x^8-y^8}+\frac{x^8-y^8}{x^8+y^8}=\frac{k^2+4}{4k}+\frac{4k}{k^2+4}$$$$=\frac{k^4+24k^2+16}{4k(k^2+4)}$$$$★if\frac{a}{b}=\frac{c}{d}then\frac{a+b}{a-b}=\frac{c+d}{c-d}$$$$componendoanddividendo$$

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