Question

If $x+\frac{1}{x}=3$, find the value of $x^3+\frac{1}{x^3}$

Answer 1

$x+\frac{1}{x}=3\Rightarrow {(x+\frac{1}{x})}^3=(3{)}^3$ ...[On cubing both sides]

$x^3+{\left(\frac{1}{x}\right)}^3+3.x.\frac{1}{x}(x+\frac{1}{x})=27$

$\Rightarrow x^3+\frac{1}{x^3}+3.3=27$ ....$[\because x+\frac{1}{x}=3]$

$\Rightarrow x^3+\frac{1}{x^3}27-8=18$

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