Question

Please tell me the answer​

Answer 1

Answer:

$\frac{ {x}^{2} + 1}{x} = 3 \frac{1}{3} = > \frac{ {x}^{2} }{x} + \frac{1}{x} = \frac{10}{3} = >$

$x + \frac{1}{x} = \frac{10}{3}$

Square on both sides, we get

${(x + \frac{1}{x} )}^{2} = \frac{100}{9} = > {x}^{2} + \frac{1}{ {x}^{2} } + 2 =$

$\frac{100}{9}$

Now subtract 4 on both sides

$x^{2} + \frac{1}{ {x}^{2} } + 2 - 4 = \frac{100}{9} - 4 = >$

${x}^{2} + \frac{1}{ {x}^{2} } - 2 = \frac{100 - 36}{9} = > {(x - \frac{1}{x}) }^{2} =$

$\frac{64}{9} = > x - \frac{1}{x} = \sqrt{ \frac{64}{9} } = > x - \frac{1}{x} = \frac{8}{3}$

We also know that

${x}^{2} + \frac{1}{ {x}^{2} } = \frac{100 - 18}{9} = > {x}^{2} + \frac{1}{ {x}^{2} } = \frac{82}{9}$

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

${x}^{3} - \frac{1}{ {x}^{3}} = ( \frac{8}{3} )( \frac{82}{9} + 1) = \frac{8 \times 91}{3 \times 9} = \frac{728}{27}$

Hope it helps you.