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Question Number 181524 by Rasheed.Sindhi last updated on 26/Nov/22

$$f(x+\frac{1}{x})=x^5+\frac{1}{x^5};f\left(3\right)=?$$$$\text{You can't use 'macro parameter character \#' in math mode}$$

Answered by Rasheed.Sindhi last updated on 26/Nov/22

$$f(x+\frac{1}{x})=x^5+\frac{1}{x^5};f\left(3\right)=?$$$$x+\frac{1}{x}=3\Rightarrow \{\begin{array}{l}x=3-\frac{1}{x}\\ \frac{1}{x}=3-x\end{array}$$$$x^5=?$$$$x=3-\frac{1}{x}$$$$\Rightarrow x^2=3x-1$$$$\Rightarrow x^3=3x^2-x=3(3x-1)-x=8x-3$$$$\Rightarrow x^4=8x^2-3x=8(3x-1)-3x=21x-8$$$$\Rightarrow x^5=21x^2-8x=21(3x-1)-8x=55x-21$$$$$$$$\frac{1}{x^5}=?$$$$\frac{1}{x}=3-x$$$$\frac{1}{x^2}=\frac{3-x}{x}=3\left(\frac{1}{x}\right)-1=3(3-x)-1=8-3x$$$$\frac{1}{x^3}=\frac{8-3x}{x}=8\left(\frac{1}{x}\right)-3=8(3-x)-3=21-8x$$$$\frac{1}{x^4}=21\left(\frac{1}{x}\right)-8=21(3-x)-8=55-21x$$$$\frac{1}{x^5}=55\left(\frac{1}{x}\right)-21=55(3-x)-21=144-55x$$$$$$$$x^5+\frac{1}{x^5}=(55x-21)+(144-55x)=123$$$$f\left(3\right)=123$$

Answered by SEKRET last updated on 26/Nov/22

$$\mathbf{x}+\frac{1}{\mathbf{x}}=3\to {\mathbf{x}}^5+\frac{1}{{\mathbf{x}}^5}=?$$$${\mathbf{x}}^2+\frac{1}{{\mathbf{x}}^2}=7{\mathbf{x}}^3+\frac{1}{{\mathbf{x}}^3}+3\cdot (\mathbf{x}+\frac{1}{\mathbf{x}})=27$$$${\mathbf{x}}^3+\frac{1}{{\mathbf{x}}^3}=27-9=18$$$$({\mathbf{x}}^2+\frac{1}{{\mathbf{x}}^2})({\mathbf{x}}^3+\frac{1}{{\mathbf{x}}^3})=7\cdot 18$$$${\mathbf{x}}^5+\frac{1}{{\mathbf{x}}^5}+\mathbf{x}+\frac{1}{\mathbf{x}}=126{\mathbf{x}}^5+\frac{1}{{\mathbf{x}}^5}=123$$$$\mathit{A}\mathit{B}\mathit{D}\mathit{U}\mathit{L}\mathit{A}\mathit{Z}\mathit{I}\mathit{Z}\mathit{A}\mathit{B}\mathit{D}\mathit{U}\mathit{V}\mathit{A}\mathit{L}\mathit{I}\mathit{Y}\mathit{E}\mathit{V}$$

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