I would suggest using some Newton-Raphson scheme.
1. Let [tex]f(x)=x^{3}-a[/tex]
You are to find X so that f(X)=0.
2. Pick an initial value [tex]x_{0}\to{f}(x_{0})=x_{0}^{3}-a[/tex]
3. The equation for the tangent line L(x)=at [tex](x_{0},f(x_{0})[/tex] is given by:
[tex]L(x)=f(x_{0})+f'(x_{0})(x-x_{0})[/tex]
4- Let the next iteration point be the x-intercept of L(x):
[tex]L(x_{1})=0\to{x}_{1}=x_{0}-\frac{f(x_{0})}{f'(x_{0}}[/tex]
5. Or, in this case, the iterative scheme becomes:
[tex]x_{n}=x_{n-1}-\frac{x_{n-1}-\frac{a}{x_{n-1}^{2}}}{3}[/tex]
That is:
[tex]x_{n}=\frac{2x_{n-1}^{3}+a}{3x_{n-1}^{2}}, n\geq{1}[/tex]