Question

If $x+\frac{1}{x}=3$ 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 $a=x,b=\frac{1}{x}$

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

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

$={\left(3\right)}^3-3\left(3\right)$

$=27-9=18$

$\therefore x^3+\frac{1}{x^3}=18$