If $P=x^3-\frac{1}{x^3}$ and $Q=x-\frac{1}{x}, x \in(0, \infty)$ then minimum value of $\frac{P}{Q^2}$, is

$2 \sqrt{3}$

We have, $P=x^3-\frac{1}{x^3}$ and $Q=x-\frac{1}{x}$

∴   $P=\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$

$\Rightarrow P=Q^3+3 Q \Rightarrow \frac{P}{Q^2}=Q+\frac{3}{Q}$

Let $Z=\frac{P}{Q^2}=Q+\frac{3}{Q}$. Then,

$\frac{d Z}{d x}=\frac{d Q}{d x}-\frac{3}{Q^2} \frac{d Q}{d x}$

For maximum/minimum, we must have

$\frac{d Z}{d x}=0$

$\Rightarrow \frac{d Q}{d x}\left\{1-\frac{3}{Q^2}\right\}=0 \Rightarrow Q^2=3 $          $\left[∵ \frac{d Q}{d x}=1+\frac{1}{x^2} \neq 0\right]$

$\Rightarrow Q= \pm \sqrt{3}$

Again, $\frac{d Z}{d x}=\left(\frac{Q^2-3}{Q^2}\right) \frac{d Q}{d x}=\frac{(1-\sqrt{3})(Q+\sqrt{3})}{Q^2} \frac{d Q}{d x}$

The changes in signs of $\frac{d Z}{d x}$ for different values of Q are shown in Figure.

Clearly, Z is minimum at $Q=\sqrt{3}$.

The minimum value of Z is given by

$Z=\frac{P}{Q^2}=\frac{3 \sqrt{3}+3 \sqrt{3}}{3}=2 \sqrt{3}$