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.

1) $n(1+1/n )^{1/3} = $
$\dfrac{(1+1/n)^{1/3}}{1/n}.$
2) $n(1+1/n^3 )^{1/3} =$
$\dfrac{(1+1/n^3)^{1/3}}{1/n}.$
$\small{\dfrac{((1+1/n)^{1/3} -1) -((1+1/n^3)^{1/3}-1)}{1/n}}$
$\small{=\dfrac{(1+1/n)^{1/3} -1}{1/n} - (1/n^2)\dfrac{(1+1/n^3)^{1/3} -1}{1/n^3}.}$
First term:
Let $f(x)=x^{1/3}$:
$\lim_{n \rightarrow \infty}\dfrac{(1+1/n)^{1/3} -1}{1/n}=$
$f'(x)_{x=1}= (1/3);$
Second term:
$\small{-\lim_{n \rightarrow \infty}(1/n^2)\dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}= }$
$\small{\lim_{n \rightarrow \infty}(1/n^2)×}$
$\small{\lim_{n \rightarrow \infty}\dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}=}$
$\small{0 \cdot (1/3)=0.}$
