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Question Number 167492 by mathlove last updated on 18/Mar/22

$$x^a=\sqrt{2}+1.........\left(1\right)$$$$x^b=\sqrt{2}-1.........\left(2\right)$$$$\frac{1}{x^{a-b}}+\frac{1}{x^{b-a}}=?$$$$\left(1\right)\mathbf{÷}\left(2\right)\Rightarrow \frac{x^a}{x^b}=\frac{\sqrt{2}+1}{\sqrt{2-1}}\Rightarrow x^{a-b}=\frac{\sqrt{2}+1}{\sqrt{2-1}}\Rightarrow \frac{1}{x^{a-b}}=\frac{\sqrt{2}-1}{\sqrt{2}+1}....\left(3\right)$$$$\left(2\right)\mathbf{÷}\left(1\right)\Rightarrow \frac{x^b}{x^a}=\frac{\sqrt{2}-1}{\sqrt{2}+1}\Rightarrow x^{b-a}=\frac{\sqrt{2}-1}{\sqrt{2}+1}\Rightarrow \frac{1}{x^{b-a}}=\frac{\sqrt{2}+1}{\sqrt{2}-1}....\left(4\right)$$$$\frac{1}{x^{a-b}}+\frac{1}{x^{b-a}}=\frac{\sqrt{2}-1}{\sqrt{2}+1}+\frac{\sqrt{2}+1}{\sqrt{2}-1}=\frac{(\sqrt{2}-1)(\sqrt{2}-1)+(\sqrt{2}+1)(\sqrt{2}+1)}{(\sqrt{2}+1)(\sqrt{2}-1)}$$$$=\frac{{(\sqrt{2}-1)}^2+{(\sqrt{2}+1)}^2}{2-1}=2-\cancel{2\sqrt{2}}+1+2+\cancel{2\sqrt{2}}+1$$$$\frac{1}{x^{a-b}}+\frac{1}{x^{b-a}}=6$$

Answered by Rasheed.Sindhi last updated on 18/Mar/22

$$x^a=\sqrt{2}+1.........\left(1\right)$$$$x^b=\sqrt{2}-1.........\left(2\right)$$$$\frac{1}{x^{a-b}}+\frac{1}{x^{b-a}}=?$$$$\frac{1}{x^{a-b}}+\frac{1}{x^{b-a}}=\frac{1}{\frac{x^a}{x^b}}+\frac{1}{\frac{x^b}{x^a}}=\frac{x^b}{x^a}+\frac{x^a}{x^b}$$$$=\frac{{\left(x^b\right)}^2+{\left(x^a\right)}^2}{x^a{x}^b}$$$$=\frac{{(\sqrt{2}-1)}^2+{(\sqrt{2}+1)}^2}{(\sqrt{2}+1)(\sqrt{2}-1)}$$$$=\frac{2{\left(\sqrt{2}\right)}^2+2{\left(1\right)}^2}{{\left(\sqrt{2}\right)}^2-{\left(1\right)}^2}$$$$=\frac{4+2}{2-1}$$$$=6$$

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