Question

if x=3-2 root2,find x cube -1/x cube

Answer 1

$x = 3 - 2 \sqrt{2} \\ \frac{ {x}^{3} - 1}{x} = \frac{ ({3 - 2 \sqrt{2}) }^{3} - 1 }{3 - 2 \sqrt{2} } = \frac{27 - 54 \sqrt{2} + 72 - 16 \sqrt{2} - 1 }{3 - 2 \sqrt{2} } = \frac{98 - 7 0\sqrt{2} }{3 - 2 \sqrt{2} } = \frac{98 - 70 \sqrt{2} }{3 - 2 \sqrt{2} } \times \frac{3 + 2 \sqrt{2} }{3 + 2 \sqrt{2 } } = \frac{(98 - 70 \sqrt{2})(3 + 2 \sqrt{2} ) }{9 - 4 \times 2} = \frac{(98 - 70 \sqrt{2} )(3 + 2 \sqrt{2} )}{1} = (98 - 70 \sqrt{2} )(3 + 2 \sqrt{2} )$
=294 + 196√2 -210√2-280
=14-14√2