Finding multivariable solutions given the limit Find $n,mER$ such that:
$limit_{x\rightarrow\infty}$   of   $(8x^3+mx^2)^{1/3}-nx$ = 1
Attempt (after multiplying by the conjugate):
$\frac{(8x^3+mx^2)-n^2x^2)}{(8x^3+mx^2)^{1/3}+nx)}$ = 1
When I reduce it down I am left with $\frac{8}{n}$= 1
I don't think this is right so am I doing something incorrectly?

Your attempt isn't quite right, since $(f^{1/3})^2 \neq f$.
Instead, since we want the growing term on the left to be compensated by the decreasing term on the left, and both grow like $x$, we can factor the first term.
$$(8x^3 +mx^2)^{1/3} = 8^{1/3} x\left(1+\frac{m}{8x}\right)^{1/3} = 8^{1/3} x\left(1+\frac{1}{3}\frac{m}{8x}+\cdots\right)$$
Where we used a binomial series. Can you finish it off now?
