Question

if [x+1/x]^2 = 3 , show that [x ^3+1/x^3]=0

Answer 1

${x}^{2} + \frac{1}{ {x}^{2} } + 2 = 3 \\ {x}^{2} + \frac{1}{ {x}^{2} } = 1 \\ ({x}^{2} + \frac{1}{ {x}^{2} })(x + \frac{1}{x} ) = 1 \times \sqrt{3} \\ {x}^{3} + \frac{1}{ {x}^{3} } + x + \frac{1}{x} = \sqrt{3} \\ {x}^{3} + \frac{1}{ {x}^{3} } + \sqrt{3} = \sqrt{3} \\ {x}^{3} + \frac{1}{ {x}^{3} } = 0$