Question

please do this question

Answer 1

Hi there !!
Here are your answers

Given,

$(x - \frac{1}{x} ) = 3$
To find the value of :

${x}^{2} + \frac{1}{ {x}^{2} }$
Using the algebraic identity (a-b)² = a²-2ab+b²,we have,


$(x - \frac{1}{x} ) {}^{2} = 3 {}^{2}$

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


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

$x {}^{2} + \frac{1}{x {}^{2} } = 9 + 2$
Thus,
${x}^{2} + \frac{1}{x {}^{2} } = 11$

Also,
the value of x⁴+1/x⁴ is also to be founded ,

In this case , the identity (a+b)² = a²+2ab+b² will be used
So,

$( {x}^{2} + \frac{1}{ {x}^{2} }) {}^{2} = 11 {}^{2}$


$({(x {}^{2}) }^{2} + 2( {x}^{2})( \frac{1}{ {x}^{2} } ) + \frac{1}{ ({ {x}^{2}) }^{2} } = 11 {}^{2}$

${x}^{4} + 2 + \frac{1}{ {x}^{4} } = 121$

${x}^{4} + \frac{1}{x {}^{4} } = 121 - 2$


$x {}^{4} + \frac{1}{x {}^{4} } = 119$
Hence the value is 119