Cubic roots and difference of cubes in limits $\lim\limits_{n\to\infty} (\sqrt[3]{n^6-6n^4+1} - n^2)$ Find the limit:
$$\lim\limits_{n\to\infty} (\sqrt[3]{n^6-6n^4+1} - n^2)$$
I applied the identity:
$$a^3 - b^3 = (a-b)(a^2+ab+b^2)$$
by multiplying the numerator and denominator by the complementary part.
$$\lim\limits_{n\to\infty} \frac{(\sqrt[3]{n^6-6n^4+1} - n^2)(\sqrt[3]{(n^6-6n^4+1)^2} + n^2\sqrt[3]{n^6-6n^4+1}+ n^4)}{(\sqrt[3]{(n^6-6n^4+1)^2} + n^2\sqrt[3]{n^6-6n^4+1}+ n^4)}=\lim\limits_{n\to\infty} \frac{n^6-6n^4+1 - n^6}{(\sqrt[3]{(n^6(1-6/n^2+1/n^6))^2} + n^2\sqrt[3]{n^6(1-6/n^2+1/n^6)}+ n^4)}=\lim\limits_{n\to\infty} \frac{-6n^4+1}{(n^4\sqrt[3]{(1-6/n^2+1/n^6)^2} + n^4\sqrt[3]{(1-6/n^2+1/n^6)}+ n^4)}=\lim\limits_{n\to\infty} \frac{-6+1/n^4}{(\sqrt[3]{(1-6/n^2+1/n^6)^2} + \sqrt[3]{(1-6/n^2+1/n^6)}+ 1)}=\frac{-6}{3}=-2$$
Is there any more elegant way to approach the problem?

With binomial series, it's $$\lim\limits_{x\to\infty}\left[ n^2\left(1-\dfrac6{n^2}+\dfrac1{n^6}\right)^{1/3}-n^2\right]$$
$$\lim\limits_{x\to\infty}\left[ n^2\left(1-\dfrac13\dfrac6{n^2}+O\left(\dfrac1{n^4}\right)\right) -n^2\right]$$
$$=-2$$
