Question

answer 15,16,17,18 with steps and write on paper

Answer 1

$15) \: x - \frac{1}{x} = \sqrt{6} \\ ( {x - \frac{1}{x}) }^{2} = ( { \sqrt{6} )}^{2} \\ {x}^{2} - 2x( \frac{1}{x} ) + \frac{1}{{x}^{2} } = 6 \\ {x}^{2} + \frac{1}{ {x}^{2} } - 2 = 6 \\ {x}^{2} + \frac{1}{ {x}^{2} } = 6 + 2 = 8 \\ {x}^{2} + \frac{1}{ {x}^{2} } = 8 \: \: \: \: ans$
16) $( {a}^{4} + \frac{1}{ {a}^{4} } ) = 727 \\ ( { {a}^{2} )}^{2} + \frac{1}{ {( {a}^{2}) }^{2} } + 2 = 727 + 2 \\ ( { {a}^{2} + \frac{1}{ {a}^{2} }) }^{2} = 729 \\ ( {a}^{2} + \frac{1}{ {a}^{2} } ) = \sqrt{729} = 27 \\ ( {a}^{2} + \frac{1}{ {a}^{2} } ) + 1 = 27 + 1 = 28 \: \: \: \: eq1\\ {a}^{2} - 2+ \frac{1}{ {a}^{2} } = 27 - 2 \\ ( {a - \frac{1}{a}) }^{2} =25 \\ (a - \frac{1}{a}) = \sqrt{25} = 5 \: \: \: \: eq2\\ now \: look \: at \: ( {a}^{3} - \frac{1}{ {a}^{3} } ) = (a - \frac{1}{a} )( {a}^{2} + a \times \frac{1}{a} + \frac{1}{ {a}^{2} } ) \\ = (a - \frac{1}{a} )( {a}^{2} + 1 + \frac{1}{ {a}^{2} } ) \: \: \: \: place \: values \: from \: eq \: 1 \: and \: eq2\\ = 5 \times 28 \\ = 140 \: \: \: ans$