Question

If $x-\frac{1}{x}$ = 2, find the value of $x^3-\frac{1}{x^3}$

Answer 1

Since$x-\frac{1}{x}=2$. Cubing both sides we get,
${(x-\frac{1}{x})}^3=2^3$
$\Rightarrow x^3-\frac{1}{x^3}-3x.\frac{1}{x}(x-\frac{1}{x})=8$
$\Rightarrow x^3-\frac{1}{x^3}-3(x-\frac{1}{x})=8$
Substituting the value of $(x-\frac{1}{x})$, we get,
$\Rightarrow x^3-\frac{1}{x^3}-3\times 2=8$
$\Rightarrow x^3-\frac{1}{x^3}-6=8$
$\Rightarrow x^3-\frac{1}{x^3}=8+6$
$\Rightarrow x^3-\frac{1}{x^3}=14$