Question

if x is equal to 9 minus 4 root 5 find then x cube + 1
upon x cube

Answer 1

$x = 9 - 4 \sqrt{5} \\ \frac{1}{x} = \frac{1}{9 - 4 \sqrt{5} } = \frac{9 + 4 \sqrt{5} }{(9 - 4 \sqrt{5} )(9 + 4 \sqrt{5} )} \\ \frac{9 + 4 \sqrt{5} }{81 - 80} = 9 + 4 \sqrt{5} \\ x + \frac{1}{x} = (9 - 4 \sqrt{5} ) + (9 + 4 \sqrt{5} ) \\ = 18 \\ {x}^{3} + \frac{1}{ {x}^{3} } = {(x + \frac{1}{x} )}^{3} - 3(x + \frac{1}{x} ) \\ = {18}^{3} - 3 \times 18 = 5778$