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Question Number 131573 by mathlove last updated on 06/Feb/21

i f f (x + \frac{1}{x}) = x^{3} + \frac{1}{x^{3}} t h e n f a i n d

f (\frac{1}{x}) = ?

Answered by rs4089 last updated on 06/Feb/21

f (x + \frac{1}{x}) = x^{3} + \frac{1}{x^{3}}

f (x + \frac{1}{x}) = (x + \frac{1}{x}) (x^{2} + \frac{1}{x^{2}} - 1)

f (x + \frac{1}{x}) = (x + \frac{1}{x}) {{(x + \frac{1}{x})}^{2} - 2 - 1}

f (x + \frac{1}{x}) = (x + \frac{1}{x}) {{(x + \frac{1}{x})}^{2} - 3}

n o w p u t x + \frac{1}{x} = t

f (t) = t (t^{2} - 3)

o r f (x) = x (x^{2} - 3)

s o f (\frac{1}{x}) = \frac{1}{x} (\frac{1}{x^{2}} - 3) = \frac{1 - 3 x^{2}}{x^{3}}

Commented by mathlove last updated on 06/Feb/21

t a n k s

Answered by mathmax by abdo last updated on 06/Feb/21

we have a^{3} + b^{3} = {(a + b)}^{3} - 3 ab (a + b) \Rightarrow

f (x + \frac{1}{x}) = {(x + \frac{1}{x})}^{3} - 3 (x + \frac{1}{x}) let x + \frac{1}{x} = t \Rightarrow f (t) = t^{3} - 3 t \Rightarrow

f (\frac{1}{x}) = \frac{1}{x^{3}} - \frac{3}{x} = \frac{x^{2} - 3}{x^{3}}