Cubic roots and difference of cubes in limits $\lim\limits_{n\to\infty} (\sqrt[3]{n^6-6n^4+1} - n^2)$

With binomial series, it's $$\lim\limits_{x\to\infty}\left[ n^2\left(1-\dfrac6{n^2}+\dfrac1{n^6}\right)^{1/3}-n^2\right]$$

$$\lim\limits_{x\to\infty}\left[ n^2\left(1-\dfrac13\dfrac6{n^2}+O\left(\dfrac1{n^4}\right)\right) -n^2\right]$$

$$=-2$$

A possible way is turning it into a derivative using $$n=\frac{1}{\sqrt t}$$

and consider $t\to 0^+$.

Hence,

$$\sqrt[3]{n^6-6n^4+1} - n^2 = \frac{\sqrt[3]{1-6t+t^3}-1}{t}$$ $$\stackrel{t\to 0^+}{\longrightarrow}\left.(\sqrt[3]{1-6t+t^3})'\right|_{t=0}=\frac{-6}{3}=-2$$