Find the limit of $\lim_{n\to\infty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$ without using the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ Find the limit of the sequence $$\lim_{n\to\infty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$$
I showed that the limit is $1/3$, using the identity $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
we get the sequence is equal to
$$\frac{n^3+n^2-n^3-1}{(n^3+n^2)^{2/3}+(n^6+n^5+n^3+n^2)^{1/3}+(n^3+1)^{2/3}}\longrightarrow\frac{1}{3}$$
but I couldn't manage to solve it without using the above identity, and I was wondering if it is possible.

$$
\begin{align}
\lim_{n\to\infty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})
&=\lim_{n\to\infty}n\left(\left(1+\frac1n\right)^{1/3}-\left(1+\frac1{n^3}\right)^{1/3}\right)\\
&=\lim_{n\to\infty}n\left(\left[1+\frac1{3n}+O\!\left(\frac1{n^2}\right)\right]-\left[1+O\!\left(\frac1{n^3}\right)\right]\right)\\
&=\lim_{n\to\infty}\left(\frac13+O\!\left(\frac1n\right)\right)\\
&=\frac13
\end{align}
$$
