Let  $f\left(x\right)=8x^3+3x$ and $f^{-1}\left(x\right)$ be the inverse function of f (x), then  $\underset{x\to \infty }{Lt}\frac{x^{\frac{1}{3}}}{f^{-1}\left(8x\right)-f^{-1}\left(x\right)}=$

$\underset{x\to \infty }{Lt}\frac{f^{-1}\left(x\right)}{x^{\frac{1}{3}}}x=f\left(t\right)$ As  $x\to \alpha ,t\to \alpha$
 $=\underset{t\to \alpha }{Lt}\frac{t}{{\left(f\left(t\right)\right)}^{\frac{1}{3}}}=\underset{t=\alpha }{Lt}\frac{t}{{\left(8t^3+3t\right)}^{\frac{1}{3}}}=\frac{1}{2}$
similarly  $\frac{f^{-1}\left(8x\right)}{x^{\frac{1}{x}}},8x=f\left(t\right)$
 $=\underset{t\to \alpha }{Lt}\frac{t}{{\left(\frac{f\left(t\right)}{8}\right)}^{\frac{1}{3}}}=2.\frac{1}{2}=1$