Question

if x+1/x=√3 then x^6 =?

Answer 1

$x + \frac{1}{x} = \sqrt{3} \\ {(x + \frac{1}{x} )}^{2} = 3 \\ {x}^{2} + \frac{1}{ {x}^{2} } = 3 - 2 = 1 - - - (1)\\ \frac{ {x}^{4} + 1}{ {x}^{2} } = 1 \\ {x}^{4} = {x}^{2} - 1 - - - (2)$
$squaring \: \: (1) \\ {( {x}^{2} + \frac{1}{{ x }^{2} } ) }^{2} = {1}^{2} \\ {x}^{4} + \frac{1}{ {x}^{4} } = 1 - 2 = - 1 \\ substituting \: \: for \: \: first \: \: {x}^{4} \\ from \: \: (2) \\ ({x}^{2} - 1) + \frac{1}{ {x}^{4} } = - 1 \\ {x}^{2} + \frac{1}{ {x}^{4} } = 0 \\ \frac{ {x}^{6} + 1 }{ {x}^{4} } = 0 \\ {x}^{6} + 1 = 0 \\ {x}^{6} = - 1$