Question

please answer correctly and step by step

*NO SPAMMING*​

Answer 1

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

Squaring Both Sides

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

Now,

$(x - \frac{1}{x} ) ^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2 \\ (x - \frac{1}{x} ) ^{2} = \frac{82}{9} - 2 \\ (x - \frac{1}{x} ) ^{2} = \frac{64}{9} \\ x - \frac{1}{x} = \frac{8}{3}$

Now We need to find Value of

${x}^{3} - \frac{1}{ {x}^{3} } = (x - \frac{1}{x} )( {x}^{2} + \frac{1}{x ^{2} } + 1) \\ {x}^{3} - \frac{1}{ {x}^{3} } = ( \frac{8}{3} )( \frac{82}{9} + 1) \\ {x}^{3} - \frac{1}{ {x}^{3} } = ( \frac{8}{3} )( \frac{91}{9} ) \\ {x}^{3} - \frac{1}{ {x}^{3} } = \frac{728}{27}$

Hope it Helps ^_^

Answer 2

Answer:

first we should find the x cube from the given and then substitute what is been asked after the calculation you get the answer