Question

x2 + 1/x2=14 find the value of x3+1/x3

Answer 1

${x}^{2} + \frac{1}{ {x}^{2} } = 14$


$add \: 2(x \times \frac{1}{x} ) \: on \: both \: sides$


${x}^{2} + \frac{1}{ {x}^{2} } + 2(x + \frac{1}{x} ) = 14 + 2(x \times \frac{1}{x} ) \\ \\ (x + \frac{1}{x})^{2} = 16 \\ \\ \\ x + \frac{1}{x} = 4$

Now,

Cube on both sides,



${(x + \frac{1}{x} )}^{3} = {4}^{3} \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3(x \times \frac{1}{x} )(x + \frac{1}{x} )= 64 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3(x + \frac{1}{x} ) = 64 \\ \\ {x}^{3} \frac{1}{ {x}^{3} } + 3(4) = 64 \\ \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 12 = 64 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = 64 - 12 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = 52$





I hope this will help you



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