f-x-1-x-x-6-1-27-f-x-

Question Number 12386 by Joel576 last updated on 21/Apr/17

f (x + \frac{1}{x}) = \frac{x^{6} + 1}{27}

f (x) = ?

Answered by ajfour last updated on 21/Apr/17

l e t (x + \frac{1}{x}) = t

s i n c e {(x + \frac{1}{x})}^{3} = x^{3} + \frac{1}{x^{3}} + 3 (x + \frac{1}{x})

t^{3} = \frac{x^{6} + 1}{x^{3}} + 3 t

\Rightarrow (t^{3} - 3 t) = \frac{(x^{6} + 1)}{\sqrt{(x^{6} + 1) - 1}}

\Rightarrow {(t^{3} - 3 t)}^{2} {(x^{6} + 1) - 1} = {(x^{6} + 1)}^{2}

{(x^{6} + 1)}^{2} - {(t^{3} - 3 t)}^{2} (x^{6} + 1) + {(t^{3} - 3 t)}^{2} = 0

(x^{6} + 1) = \frac{{(t^{3} - 3 t)}^{2} \pm \sqrt{{(t^{3} - 3 t)}^{4} - 4 {(t^{3} - 3 t)}^{2}}}{2}

f (t) = \frac{(x^{6} + 1)}{27} = \frac{{(t^{3} - 3 t)}^{2} \pm \sqrt{{(t^{3} - 3 t)}^{4} - 4 {(t^{3} - 3 t)}^{2}}}{54}

f (x) = \frac{{(x^{3} - 3 x)}^{2}}{54} [1 \pm \sqrt{1 - \frac{4}{{(x^{3} - 3 x)}^{2}}}] .

Commented by ajfour last updated on 21/Apr/17

f o r x = 1

f (1 + 1) = \frac{1 + 1}{27} = \frac{2}{27} .

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/Apr/17

x + x^{- 1} = t \Rightarrow x^{2} - t x + 1 = 0 \Rightarrow x = \frac{t \pm \sqrt{t^{2} - 4}}{2}

x^{2} = t x - 1

x^{3} = {t x}^{2} - x = t (t x - 1) - x = (t^{2} - 1) x - t

x^{4} = (t^{2} - 1) x^{2} - t x = (t^{2} - 1) (t x - 1) - t x =

= (t^{2} - 2) t x - (t^{2} - 1)

x^{6} = x^{4} \times x^{2} = [(t^{2} - 2) t x - (t^{2} - 1)] (t x - 1) =

= (t^{2} - 2) t^{2} x^{2} - (t^{2} - 2) t x - (t^{2} - 1) t x + (t^{2} - 1) =

= t^{2} (t^{2} - 2) (t x - 1) - [(t^{2} - 2 + t^{2} - 1)] t x + (t^{2} - 1) =

(t^{4} - 2 t^{2} - 2 t^{2} + 3) t x - (t^{4} - 2 t^{2} - t^{2} + 1) \Rightarrow

x^{6} = \frac{1}{2} t (t^{4} - 4 t^{2} + 3) (t \pm \sqrt{t^{2} - 4}) - (t^{4} - 3 t^{2} + 1)

x^{6} + 1 = \frac{1}{2} t (t^{2} - 3) [(t^{2} - 1) (t \pm \sqrt{t^{2} - 4}) - 2 t]

f (x) = \frac{x (x^{2} - 3)}{54} [(x^{2} - 1) (x \pm \sqrt{x^{2} - 4}) - 2 x]