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Answer» GIVEN :- ${<klux>X</klux>}^{2} + \dfrac{1}{ {x}^{2} } = 18$ To find :- $x - \dfrac{1}{x}$ Identity to USE :- (a-b)² = a² - 2ab + b² substituting the values, $(x - \dfrac{1}{x})^{2} = ({x})^{2} - 2(x)( \dfrac{1}{x}) + (\dfrac{1}{x})^{2}$ $(x - \dfrac{1}{x})^{2} = {x}^{2} - 2( \not x)( \dfrac{1}{ \not x}) + \dfrac{1}{ {x}^{2} }$ $(x - \dfrac{1}{x})^{2} = {x}^{2} - 2 + \dfrac{1}{ {x}^{2} }$ By rearranging, $(x - \dfrac{1}{x})^{2} = {x}^{2} + \dfrac{1}{ {x}^{2} } - 2$ By substituting the VALUE for x²+1/x², $(x - \dfrac{1}{x})^{2} = 18 - 2$ $(x - \dfrac{1}{x})^{2} = 16$ $(x - \dfrac{1}{x}) = √16$ $(x - \dfrac{1}{x}) = 4$ Some more identities : (a+b)² = a² + 2ab + b² a²-b² = (a-b)(a+b) (x+a)(x+b) = x² + x(a+b) + ab (a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca

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