Question

If $x-\frac{1}{x}=5$ then find the value of $x^3-\frac{1}{x^3}$

Answer 1

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

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

Take $a=x,b=\frac{1}{x}$

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

$x^3-\frac{1}{x^3}={\left(5\right)}^3+3\left(5\right)$

$x^3-\frac{1}{x^3}=125+15=140$

$\therefore x^3-\frac{1}{x^3}=140$