Question

factors of x^2+1/x^2-3 are

Answer 1

$Answer: \\ \\ Now, \: \: {x }^{2} + \frac{1}{ {x}^{2} } -3 \\ \\ = {(x + \frac{1}{x}) }^{2} - (2 \times x \times \frac{1}{x} ) - 3 \\ \\ = {(x + \frac{1}{x}) }^{2} - 2 - 3 \\ \\ = {(x + \frac{1}{x} )}^{2} - 5 \\ \\ = {( x + \frac{1}{x}) }^{2} - {( \sqrt{5}) }^{2} \\ \\ = (x + \frac{1}{x} + \sqrt{5} )(x + \frac{1}{x} - \sqrt{5} ), \\ using \: \: the \: \: identity \\ (x + y)(x - y) = {x}^{2} - {y}^{2} \\ \\ which \: \: is \: \: the \: \: required \\ factorization.$

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Answer 2

Solution:-

given by:-

${x}^{2} + \frac{1}{ {x}^{2} } - 3 \\ by \: the \: for \: formula \\ {(x + \frac{1}{x} )}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2 \\ {(x + \frac{1}{x} ) }^{2} - 5 = {x}^{2} + \frac{1}{ {x}^{2} } - 3 \\ {(x + \frac{1}{x} ) }^{2} - { \sqrt{5}}^{2} \\ condition \: aply \: by \: diffrence \: square \: method \\ (x + \frac{1}{x} + \sqrt{5} )(x + \frac{1}{x} - \sqrt{5} )ans$