Question

find the value of
$x {3 + 1 \div x3}^{?}$

Answer 1

Hi ,

x = 3 - √8

x = 3 - √8 --( 1 )

1/x = 1/( 3 - √8 )

= ( 3 + √8 )/[ ( 3 - √8 )( 3 + √8 ) ]

= ( 3 + √8 )/ [ 3² - (√8 )² ]

= ( 3 + √8 )/ ( 9 - 8 )

= 3 + √8 ----( 1 )

Now ,

x³ + 1/x³

= ( x + 1/x )³ - 3( x + 1/x )


= [ 3 - √8 + 3 + √8 ]³ - 3[ 3 -√8 + 3 + √8 ]

= 6³ - 3 × 6

= 6 ( 6² - 3 )

= 6( 36 - 3 )

= 6 × 33

= 198

I hope this helps you.

: )

Answer 2

Heya !!

Here's your answer.. ⬇⬇
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$x = 3 - \sqrt{8} \\ \\ \frac{1}{x} = \frac{1}{3 - \sqrt{8} } \\ \\ \frac{1}{x} = \frac{1}{3 - \sqrt{8} } \times \frac{3 + \sqrt{8} }{3 + \sqrt{8} } \\ \\ \frac{1}{x} = \frac{3 + \sqrt{8} }{9 - 8 } \\ \\ x + \frac{1}{x} = 3 - \sqrt{8} + 3 + \sqrt{8} \\ \\ x + \frac{1}{x} = 6 \\ \\ take \: \: cube \: \: on \: \: both \: \: side \: \: \\ \\ {(x + \frac{1}{x} )}^{3} = {(6)}^{3} \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3x \times \frac{1}{x} (x + \frac{1}{x} ) = 216 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3(6) = 216 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = 216 - 18 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = 198$
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Hope it helps..

Thanks :)