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

i f x \neq 0 t h e n

p r o v e \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}

c o n t i n u e

Commented by Nayon last updated on 28/Sep/16

w r o n g w a y f i l u p

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

S e e a n s w e r i n t h e f o l l o w i n g c o m m e n t b y n u m e 1114 .

Commented by Nayon last updated on 28/Sep/16

t h e q u e s t i o n i s t h a t t o p r o v e i t . {i t}^{'} s a i d e n t i t y,

n o t o n l y a e q u e t i o n .

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

S o r r y I m i s r e a d .

Commented by nume1114 last updated on 28/Sep/16

L e t x + x^{- 1} = y

x^{2} + x^{- 2} = y^{2} - 2

x^{3} + x^{- 3} = y^{3} - 3 y

x^{4} + x^{- 4} = y^{4} - 4 y^{2} + 2

L H S = \frac{y^{4} - 4 y^{2} + 3}{y^{3} - 3 y} = \frac{(y^{2} - 3) (y^{2} - 1)}{y (y^{2} - 3)}

[\begin{matrix} i f x > 0 \\ y = x + x^{- 1} ⩾ 2 \sqrt{{x x}^{- 1}} = 2 \\ i f x < 0 \\ - y = - x + {(- x)}^{- 1} ⩾ 2 \sqrt{(- x) \cdot (- x^{- 1})} = 2 \\ y ⩽ - 2 \\ s o, y^{2} - 3 \neq 0 \end{matrix}]

\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}} = R H S

P r o v e d

Answered by sandy_suhendra last updated on 28/Sep/16

L H S = \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} (p r o v e d)