Question

if
${x }^{2} + 1 \div {x}^{2} = 34$
then find the value of
${x}^{3} + 1 \div {x}^{3}$
​

Answer 1

${x}^{2} + \frac{1}{ {x}^{2} } = 34 \\ \\ {(x + \frac{1}{x})} ^{2} - 2 \times x + \frac{1}{x} = 34 \\ \\ {(x + \frac{1}{x} }^{2} ) - 2 = 34 \\ \\ {(x + \frac{1}{x} }^{2} ) = 34 + 2 = 36 \\ \\ \sqrt{ {(x + \frac{1}{x} }^{2} } ) = \sqrt{36} \\ \\ x + \frac{1}{x} = 6$

Then,

${x}^{3} + \frac{1}{ {x}^{3} } = {(x + \frac{1}{x}) }^{3} - 3 \times x \times \frac{1}{x} \times x + \frac{1}{x} \\ \\ {x + \frac{1}{x} }^{3} - 3(x + \frac{1}{x} ) \\ \\ {6}^{3} - 3(6) \\ \\ 108 - 18 = 90$