Question

If ${(x+\frac{1}{x})}^2-2=7$. Find the value of $x^3+\frac{1}{x^3}$.

Answer 1

$\Rightarrow {(x+\frac{1}{x})}^2-2=7$

$\Rightarrow {(x+\frac{1}{x})}^2=9$

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

Cube on both sides

$\Rightarrow x^3+\frac{1}{x^3}+3x\frac{1}{x}(x+\frac{1}{x})=27$

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

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