Question

If $x=\frac{1}{x}=-2$ then find $x^3+\frac{1}{x^3}$

Answer 1

$x+\frac{1}{x}=-2$

$x^3+\frac{1}{x^3}={\left(x+\frac{1}{x}\right)}^3-3x.\frac{1}{x}(x+\frac{1}{x})$

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

$x^3+\frac{1}{x^3}={(-2)}^3-\mathrm{3.1.}(-2)$

$=-8+6$

$=-2$