Question

x square + 1/x square. find the value of x cube +1/x cube​

Answer 1

let say

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

now to find

${x}^{3} + \frac{1}{ {x}^{3} }$

as we know that

${a}^{3} + {b}^{3} = (a + b)^3 - 3ab(a + b)$

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

$( {x + \frac{1}{x} })^{2} = {x}^{2} + { \frac{1}{x} }^{2} + 2 \\ \\ ( {x + \frac{1}{x} })^{2} = y + 2 \\ \\ x + \frac{1}{x} = \pm \sqrt{y + 2}$

putting this value we can evaluate

${x}^{3} + \frac{1}{{x}^{3} }$

$\pm \sqrt{y + 2} \times (y + 2 - 3) \\ \\ \pm \sqrt{y + 2} \times (y - 1)$