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Question Number 8016 by Nayon last updated on 28/Sep/16

$$$$$$$$$$ifx\ne 0then$$$$prove\to$$$$\frac{x^4+x^{-4}+1}{x^3+x^{-3}}=\frac{x^2+1}{x}-\frac{x}{x^2+1}$$

Commented by FilupSmith last updated on 28/Sep/16

$$\frac{x^4+\frac{1}{x^4}+1}{x^3+\frac{1}{x^3}}=\frac{\frac{x^8+1+x^4}{x^4}}{\frac{x^6+1}{x^3}}$$$$=\frac{x^8+x^4+1}{x^4}+\frac{x^3}{x^6+1}$$$$=x^4+1+\frac{1}{x^4}+\frac{x^3}{x^6+1}$$$$continue$$

Commented by Nayon last updated on 28/Sep/16

$$wrongwayfilup$$

Answered by Rasheed Soomro last updated on 28/Sep/16

$$Seeanswerinthefollowingcommentbynume1114.$$

Commented by Nayon last updated on 28/Sep/16

$$thequestionisthattoproveit.{it}^{\prime }saidentity,$$$$notonlyaequetion.$$

Commented by Rasheed Soomro last updated on 28/Sep/16

$$SorryImisread.$$

Commented by nume1114 last updated on 28/Sep/16

$$Letx+x^{-1}=y$$$$x^2+x^{-2}=y^2-2$$$$x^3+x^{-3}=y^3-3y$$$$x^4+x^{-4}=y^4-4y^2+2$$$$LHS=\frac{y^4-4y^2+3}{y^3-3y}=\frac{(y^2-3)(y^2-1)}{y(y^2-3)}$$$$\left[\begin{array}{c}ifx>0\\ y=x+x^{-1}⩾2\sqrt{{xx}^{-1}}=2\\ ifx<0\\ -y=-x+{(-x)}^{-1}⩾2\sqrt{(-x)\cdot (-x^{-1})}=2\\ y⩽-2\\ so,y^2-3\ne 0\end{array}\right]$$$$\frac{(y^2-3)(y^2-1)}{y(y^2-3)}=\frac{y^2-1}{y}=y-\frac{1}{y}$$$$=x+\frac{1}{x}-\frac{1}{x+\frac{1}{x}}=RHS$$$$Proved$$

Answered by sandy_suhendra last updated on 28/Sep/16

$$LHS=\frac{x^4+x^{-4}+1}{x^3+x^{-3}}\times \frac{x^4}{x^4}$$$$=\frac{x^8+x^4+1}{x^7+x}$$$$=\frac{{(x^4+1)}^2-x^4}{x(x^6+1)}$$$$=\frac{(x^4+1-x^2)(x^4+1+x^2)}{x(x^2+1)(x^4-x^2+1)}$$$$=\frac{x^4+x^2+1}{x(x^2+1)}$$$$=\frac{{(x^2+1)}^2-x^2}{x(x^2+1)}$$$$=\frac{{(x^2+1)}^2}{x(x^2+1)}-\frac{x^2}{x(x^2+1)}$$$$=\frac{x^2+1}{x}-\frac{x}{x^2+1}\left(proved\right)$$

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