Question

ayuda por favor $Si: x^2+1=4x\:\:clacular\:x^3+x^{-3}$

Answer 1

Recuerda:

${(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)$

...

$\frac{ {x}^{2} + 1 }{x} = \frac{4x}{x}$

$x + \frac{1}{x} = 4$

${(x + \frac{1}{x}) }^{3} = {4}^{3}$

${x}^{3} + \frac{1}{ {x}^{3} } + 3(x)( \frac{1}{x} )(x + \frac{1}{x} ) = 64$

${x}^{3} + {x}^{ - 3} + 3(1)(4) = 64$

${x}^{3} + {x}^{ - 3} = 52$