please answer correctly and step by step
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please answer correctly and step by step
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[tex]x + \frac{1}{x} = \frac{10}{3} [/tex]
Squaring Both Sides
[tex] {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} [/tex]
Now,
[tex](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} [/tex]
Now We need to find Value of
[tex] {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} [/tex]
Hope it Helps ^_^
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