The real parameters m and n are such that the graph of the function $f\left(x\right)=\sqrt[3]{8x^3+m{x}^2}-nx$  has the horizontal asymptote $y=1$ then 

We supposed to find m and n such that  $\underset{x\to \infty }{\lim }\text{3}\sqrt[\text{3}]{{\text{8x}}^{\text{3}}{\text{+mx}}^{\text{2}}}\text{-nx=1}$ or
 $\underset{x\to \infty }{\lim }\text{3}\sqrt[\text{3}]{{\text{8x}}^{\text{3}}\text{+mx}}\text{-nx=1.}$
We compute
 $\sqrt[\text{3}]{{\text{8x}}^{\text{3}}{\text{+mx}}^2}\text{-nx=}\frac{(8-n^3)x^3+m{x}^2}{\sqrt[3]{{(8x^2+m{x}^2)}^2+nx\sqrt[3]{8x^3+m{x}^2}+n^2{x}^2}}.$
$8-n^3$  must be equal to 0
N = 2
Now  $f\left(x\right)=\frac{m}{\sqrt[3]{{(8+\frac{3}{x})}^2+2\sqrt[3]{8+\frac{m}{x}+4}}}.$ 
We see that $\underset{x\to \infty }{\lim }f\left(x\right)==\frac{m}{12}.$  For this to be equal to 1, m must be equal to 12. Hence the answer to the problem is (m, n) = (12, 2).