Question

If $x-\frac{2}{x}=3$, then find $x^3-\frac{8}{x^3}$.

Answer 1

Given: $x-\frac{2}{x}=3\dots \left(i\right)$

We have to find $x^3-\frac{8}{x^3}$

Cube on both side of equation $\left(i\right),$

${(x-\frac{2}{x})}^3=(3{)}^3$

$\Rightarrow x^3-\frac{2^3}{x^3}-3\times x\times \left(\frac{2}{x}\right)(x-\frac{2}{x})=27$ $[$by using $(a-b{)}^3=a^3-b^3-3ab(a-b\left)\right]$

$\Rightarrow x^3-\frac{8}{x^3}-6\left(3\right)=27$

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

$\Rightarrow x^3-\frac{8}{x^3}=45$