Question

if x+1/x=9,then find the value of( i) X square +1/x square (ii) X cube + 1/x cube

Answer 1

Hello friends!!

If
$x + \frac{1}{x } = 9$
Then

$i) {x}^{2} + \frac{1}{ {x}^{2} }$


$x + \frac{1}{x} = 9$

Squaring both sides,

${(x + \frac{1}{x}) }^{2} = {(9)}^{2}$


Using identity : ( a + b)² = a² + b² + 2ab


${x}^{2} + \frac{1}{ {x}^{2} } + 2 \times x \times \frac{1}{x} = 81$


${x}^{2} + \frac{1}{ {x}^{2} } + 2 = 81$


${x}^{2} + \frac{1}{ {x}^{2} } = 81 - 2$



${x}^{2} + \frac{1}{ {x}^{2} } = 79$

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$ii) {x}^{3} + \frac{1}{ {x}^{3} }$


$x + \frac{1}{x} = 9$


Cube both sides

${(x + \frac{1}{x} )}^{3} = {(9)}^{3}$

Using identity : ( a + b)³ = a³ + b³ + 3ab ( a + b )


${x}^{3} + \frac{1}{ {x}^{3} } + 3 \times x \times \frac{1}{x} (x + \frac{1}{x}) = 729$


${x}^{3} + \frac{1}{ {x}^{3} } + 3(9) = 729$



${x}^{3} + \frac{1}{ {x}^{3} } + 27 = 729$

${x}^{3} + \frac{1}{ {x}^{3} } = 729 - 27$


${x}^{3} + \frac{1}{ {x}^{3} } = 702$



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HOPE IT HELPS YOU..

Answer 2

x+1/x=9
squaring on both sides
x square+1/x square+2*x*1/x=81
x square +1/x square=81-2=79

now second one
same as the first
instead of squaring do cubing
=>x cube + 1/x cube +3x*1/x(x+1/x)=729
=>x cube + 1/x cube + 3(9)=729
=>x cube+1/x cube=729-27=702