If x+1/x=4,then find the value of x^2[x^2+1/x^6]+[x^3+1/x^3].

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If x + (1/x) = 3, find x² + (1/x²) and x³ + (1/x³)

\frac{x}{1 - x^{2}} - \frac{x^{3}}{1 - x^{6}} + \frac{x^{5}}{1 - x^{10}} - . . . . . . . = \frac{x}{1 + x^{2}} + \frac{x^{3}}{1 + x^{6}} + \frac{x^{5}}{1 + x^{10}} + . . . . .

x^{2} + \frac{1}{x^{2}} = 6

x^{3} - \frac{1}{x^{3}}

x^{6} + \frac{1}{x^{6}}

{sin}^{- 1}

[x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - . . . . . . . \infty]

{cos}^{- 1}

[x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - . . . . . . . \infty]

\frac{π}{2}

\leq \sqrt{2}

x =

{sin}^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - . . . . \infty) + {cos}^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - . . . . \infty) = \frac{π}{2}

0 < x < \sqrt{2}

x

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