Question

if x equals to 3 + under root 8 find x cube + 1 upon x cube

Answer 1

x = 3 + √8

$\frac{1}{x} = \frac{1}{3 + \sqrt{8} }$

By Rationalization, we get,


$\frac{1}{x} = \frac{1}{3 + \sqrt{8} } \: \times \frac{3 - \sqrt{8} }{3 - \sqrt{8} } \\ \\ \\ = > \frac{1}{x} = \frac{3 - \sqrt{8} }{(3 + \sqrt{8} )(3 - \sqrt{8)} }$



$\mathbf{we \: know \: \: (a - b)(a + b) = {a}^{2} - {b}^{2} }$

So,


$=> \frac{1}{x} = \frac{3 - \sqrt{8} }{ {3}^{2} - {( \sqrt{8}) }^{2} } \\ \\ = > \frac{1}{x} = \frac{ 3 - \sqrt{8} }{9 - 8} = 3 - \sqrt{8}$



Hence,


$x + \frac{1}{x} \\ \\ = > 3 + \sqrt{8} + 3 - \sqrt{8 } \\ \\ => 6 \\ \\ \\ x + \frac{1}{x} = 6 \\ \\ \\ \mathbf{cube \: \: on \: \: both \: \: sides \: \: } \\ \\ \\ = > {(x + \frac{1}{x} )}^{3} = {6}^{3} \\ \\ \\ => {x}^{3} + \frac{ 1 }{ {x}^{3} } + 3(x + \frac{1}{x} )(x \times \frac{1}{x} ) = 216 \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } + 3(1)(6) = 216 \\ \\ = > {x}^{ 3} + \frac{1}{ {x }^{3} } = 216 - 18 \\ \\ => {x}^{3} + \frac{1}{ {x}^{3} } = 198$

Answer 2

Step-by-step explanation:

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