a.
[tex]\dfrac{n^{1/3}}{1-n^{1/3}}=\dfrac1{\frac1{n^{1/3}}-1}[/tex]
As [tex]n\to\infty[/tex], the [tex]n^{-1/3}[/tex] term will converge to 0, so [tex]a_n\to-1[/tex].
b. If you mean
[tex]n^{1/3}-(n^3-1)^{1/3}[/tex]
then the sequence diverges, since [tex](n^3-1)^{1/3}[/tex] behaves like [tex]n[/tex], and [tex]n>n^{1/3}[/tex] for [tex]n>1[/tex].
But if you mean
[tex]n^{1/3}-(n-1)^{1/3}[/tex]
rewrite as
[tex]\dfrac{\left(n^{1/3}-(n-1)^{1/3}\right)\left(n^{2/3}+n^{1/3}(n-1)^{1/3}+(n-1)^{2/3}\right)}{n^{2/3}+n^{1/3}(n-1)^{1/3}+(n-1)^{2/3}}=\dfrac{n-(n-1)}{n^{2/3}+n^{1/3}(n-1)^{1/3}+(n-1)^{2/3}}=\dfrac1{n^{2/3}+n^{1/3}(n-1)^{1/3}+(n-1)^{2/3}}[/tex]
which converges to 0 as [tex]n\to\infty[/tex].
